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PEACTICAL  TREATISE 


ON 


ALaEBRA, 


DESIGNED    FOR   THE   USE    OF   STUDENTS 


IN 


HIGH  SCHOOLS  AND  ACADEMIES. 


BY  BENJAMIN  GREENLEAF,  A.M., 

Arxnou  OF  toe  "  national  arithmetic,"  ktc. 


Si'irt^  Kmprobeti  Stereotype  23oitiou. 


BOSTON: 
PUBLISHED   BY  ROBERT  S.  DAVIS   k   CO, 

NEW  YORK  :  Geo.  F.  Cooledge  &  Brother,  and  D.  Bcrgess  &  Co. 

PHILADELPHIA:   Lippincott,  Grambo  &  Co. 

ST.  LOUIS  :   John  Hals  all. 

1854. 


Entered  according  to  Act  of  Congress,  in  the  year  1852,  by 

BENJAMIN    GREENLEAF, 

In  the  Clerk's  OfEce  of  the  District  Court  cf  the  District  of  Massachusetts. 


Entered  according  to  Act  of  Congress,  in  the  year  1853,  by 

BENJAMIN    GREENLEAE, 

In  the  Clerk's  Office  of  the  District  Court  of  the  District  of  Massachusetts. 


GREENLEAF'S   SERIES    OF    MATHEMATICS. 

1.  MENTAL   ARITHMETIC,  upon  the  Inductive  Plan;   designed  for  Primary  and 
Intermediate  Schools.    Revised  and  enlarged  edition.     144  pp. 

2.  INTRODUCTION  TO  THE  NATIONAL  ARITHMETIC  ;   Ok,  COMMON  SCHOOL 
ARITHMETIC.    Improved  stereotype  edition.    324  pp. 

3.  THE    NATIONAL    ARITHMETIC,    designed    for  advanced    scholars  in  Common 
Schools  and  Academies.    Improved  stereotype  edition.    360  pp. 

COMPLETE   KEYS  TO  THE    INTRODUCTION  AND    NATIONAL   ARITHMETIC, 

containing  Solutions  and  Explanations,  for  Teachers  only.    2  vols. 

4.  PRACTICAL  TREATISE  ON   ALGEBRA,  for  High  Schools  and  Academies,  and  for 
advanced  Students  in  Common  Schools.    New  edition,  revised  and  stereotyped. 

KEY  TO  THE  PRACTICAL  ALGEBRA,  containmg  the  Answers,  and  full  Solutions 
and  Explanations,  for  Teachers  only.     Stereotype  edition.     278  pp. 


EDUCATION  LIBBf 


stereotyped  by 

nOBART   &    IlOBBIJSrS, 

nrrw  England  type  and  stereotype  fodndert, 

R  O  ST  o  N. 


PREFACE. 


The  following  Treatise  is  designed  to  present  a  system  of 
theoretical  and  practical  Algebra.  It  is  intended  to  be  both 
elementary  and  comprehensive,  and  adapted  to  the  wants  of 
beginners,  as  well  as  those  who  are  advanced  in  the  study. 

In  the  course  of  his  labors  the  author  has  consulted  the  most 
approved  European  treatises  on  the  subject,  and  availed  himself 
of  whatever  he  thought  might  add  to  the  interest  and  usefulness 
of  his  work. 

It  has  been  the  aim  of  the  author,  throughout  his  investiga- 
tions, to  give  to  it  a  practical  character,  that  those  who  study  it 
may  know  how  to  apply  their  knowledge  to  useful  purposes. 

The  demonstrations  connected  with  the  several  Roots,  will 
greatly  aid  those  who  wish  for  a  complete  and  thorough  knowl- 
edge of  Evolution  in  Arithmetic. 

The  method  of  solving  Cubic  Equations  by  completing  the 
square,  the  author  believes,  will  be  very  useful.  This  method 
will  not  apply  to  all  problems ;  but,  wherever  it  will  apply,  it 
not  only  very  much  abridges  the  labor,  but  the  result  is  perfect 
accuracy,  which  is  not  always  the  fact  by  the  common  method 
of  approximation.  The  Table  of  Logarithms  at  the  end  of  the 
work,  will  be  often  found  convenient  and  useful. 

The  examples,  of  which  a  large  number  have  been  placed 
under  each  Kule,  are  intended  to  be  neither  too  numerous  nor 
too  difficult;  and  all  who  may  use  the  work,  either  by  themselves 
or  in  connection  with  a  class,  are  advised  to  solve  all  the  prob- 
lems, in  the  order  in  which  they  are  given.  No  labor  on  the 
part  of  the  pupil  will  be  productive  of  more  intellectual  and 
practical  benefit.  The  answers  to  several  questions  have  been 
designedly  omitted,  that  the  pupil  may  try  his  skill  as  upon  an 
original  problem. 


IV  PREFACE. 

One  who  has  a  thorough  knowledge  of  Arithmetic,  will  find 
the  study  of  Algebra  a  most  pleasing,  and,  generally,  not  a 
difficult  task.  As  a  mental  exercise,  it  is  admirable  for  its  efiect 
upon  the  logical  powers  of  the  mind,  assisting  one  to  think  and 
reason  closely  and  conclusively.  As  Mr.  Locke  has  remarked, 
in  his  Essay  on  the  Human  Understanding,  "Nothing  teaches  a 
man  to  reason  so  well  as  Mathematics,  which  should  be  taught 
to  all  those  who  have  time  and  opportunity,  not  so  much  to 
make  them  mathematicians,  as  to  make  them  reasonable  crea- 
tures  " 

BENJAMIN  GREENLEAP. 
Bradford,  January  23, 1852. 


ADVERTISEMENT  TO  THE  STEREOTYPE  EDITION. 

In  revising  this  work  for  a  second  edition,  the  author  has  made  such 
changes  and  additions  as  he  believed  would  better  adapt  it  to  its  purpose. 
Every  part  of  it  has  been  carefully  and  critically  examined,  and  many 
portions  have  been  entirely  re-written.  In  a  few  cases,  where  improve- 
ment in  that  respect  seemed  desirable,  the  arrangement  of  articles  has 
been  somewhat  altered. 

The  new  articles  which  have  been  inserted,  will,  it  is  hoped,  add  ma- 
terially to  the  interest,  as  well  as  to  the  value,  of  the  treatise.  The  theory 
of  Equations  has  been  more  fully  developed,  and  illustrated  by  a  variety  of 
carefully  prepared  examples.  A  brief  space  has  been  given  to  Indeter- 
minate Analysis,  a  subject  which,  though  usually  omitted  in  elementary 
works  on  Algebra,  the  author  believes  to  be  one  of  no  small  practical 
importance.  It  gives  the  student  the  command  of  a  class  of  problems 
which  cannot  possibly  be  solved  by  the  rules  of  Arithmetic,  nor  by  the 
more  familiar  principles  of  Algebra. 

In  the  revision  of  the  work,  the  author  has  availed  himself  of  the 
suggestions  of  several  teachers  who  have  used  it  as  a  text-book  since  its 
first  publication  ;  and  he  would  take  this  opportunity  to  express  his 
gratitude  for  their  kindness. 

April  26,  1853. 


CONTENTS. 


SECTION  I. 

PAGE 

Definitions  and  Notations 9 

Explanation  of  the  Signs -i*;- .      10 — 15 

Practical  Examples 15 

SECTION  II. 

Addition 16 

SECTION  III. 

Subtraction 21 

Simple  Quantities 22 

Compound  Quantities 28 

SECTION  IV. 

Multiplication 28 

Multiplication  by  detached  Coefficients 35 

SECTION  V. 

Division 37 

Division  by  detached  Coefficients , 43 

Questions  to  exercise  the  foregoing  Rules 45 

SECTION  VI. 

Fractions 46 

Definitions  of  different  Fractions 46,  47 

To  find  the  greatest  Common  Measure  of  the  terms  of  a  Fraction    .  48 

To  reduce  Fractions  to  their  lowest  terms 50 

To  reduce  a  Mixed  Quantity  to  the  form  of  a  Fraction 52 

To  represent  a  Fraction  in  the  form  of  a  Whole  or  Mixed  Quantity  53 

To  reduce  a  Complex  Fraction  to  a  Simple  one 54 

To  reduce  Fractions  to  a  common  Denominator 56 

Addition  of  Fractions 59 

To  add  Fractional  Quantities 59 

Subtraction  of  Fractions 61 

1# 


VI  CONTENTS. 

PAGB 

To  subtract  one  Fraction  from  another 61 

Multiplication  of  Fractions 63 

To  multiply  Fractions  together 63 

Division  of  Fractions 65 

To  divide  one  Fraction  by  another 65 

Negative  Exponents 66 

To  free  Fractions  from  Negative  Exponents 67 

SECTION  vn. 

Equations 68 

SECTION  VIII. 

Problems 78 


SECTION  IX. 
Equations    or    the    Fikst   Degree,   containing  two   Unknown 

Quantities 89 

Elimination  by  Addition  and  Subtraction 89 

Elimination  by  Comparison      91 

Elimination  by  Substitution 92 

SECTION  X. 

ELI>nNATION  WHERE    THERE  ARE   THREE  OR  MORE  UNKNOWN  TeRMS 

INVOLVED   IN  AN   EQUAL  NUMBER  OF  EQUATIONS 95 

Equations  or  the  First  Degree,  containing  several  Unknown 

Quantities 98 

SECTION  XI. 

Negative  Quantities 108 

Problem  of "  The  Couriers  " •  .  106 

Theory  of  Indetermination Ill 

SECTION  XII. 
Theorems  in  Negative  Quantities     .   .   .  *. 113 

SECTION  XIII. 
Involution 121 

SECTION  XIV. 

Evolution,  or  the  Extraction  of  Koots 125 

Evolution  of  Polynomials 126 

Evolution  by  Detached  Coefficients 130 

Extraction  of  the  Square  Root  of  Numbers 131 

Cube  PvOOt 132 

Cube  Root  by  Detached  Coefficients 134 


CONTENTS.  VII 

SECTION  XV. 

PAGE 

Surds,  or  Kadical  Quantities 138 

To  EXTRACT  THE  SqUARE  RoOT  OF  A  BiNOMIAL  SuRD 160 

SECTION   XVI. 

IlMAGINARY   QUANTITIES 162 

SECTION  XVII. 

Quadratic  Equations,  or  Equations  op  the  Second  Degree  .   .  164 

The  Theory  of  the  Lights  and  Attraction 168 

Application  of  the  above  Principles 169 

Affected  Quadratic  Equations 172 

Problems  in  Quadratic  Equations 186 

Examples  of  one  or  more  Unknown  Terms  in  Quadratic  Equations  192 

SECTION  XVIII. 

Cubic  and  Higher  Equations 198 

SECTION   XIX. 

Ratios 202 

Comparison  of  Ratios 203 

Compound  Ratios 204 

Proportion 205 

Approximation  of  Ratios 207 

Problems  for  Proportion 216 

SECTION  XX. 

Arithjietical  Progression 218 

Table  of  twenty  different  Cases  in  Arithmetical  Progression  .   .   .  231 

SECTION  XXI. 

Geometrical  Progression,  or  Progression  by  Quotient    .   .   .  282 

Table  of  twenty  different  Cases  in  Geometrical  Progression    .   .    .  240 

SECTION  XXII. 

Harmonical  Progression 241 

SECTION  xxni. 

Infinite  Series 242 

SECTION  XXIV. 

SiBiPLE  Interest 244 

SECTION  XXV. 

Discount  at  SniPLE  Interest 248 


VIII  CONTENTS. 

SECTION  XXVI. 

PAGE 

Partnership,  or  Company  Business 250 

Partnership  on  Time,  or  Double  Fellowship  . 253 

SECTION  xxvn. 

Indeterminate  Analysis 257 

SECTION  XXVIII. 
Variations,  Permutations,  and  Combinations 266 

SECTION  XXIX. 

Logarithms 270 

Multiplication  of  Logarithms 281 

Division  by  Logarithms 282 

Involution  by  Logarithms 286 

Evolution  by  Logarithms 286 

SECTION  XXX. 
Compound  Interest     .    • 289 

Discount  and  Present  Value  at  Compound  Interest 294 

SECTION  XXXI. 
Deposits 295 

SECTION  XXXII. 
Exponential  or  Transcendental  Equations 299 

SECTION  xxxin. 

Annuities 803 

SECTION  XXXIV. 
Involution  op  Binomials 807 

SECTION  XXXV. 

Binomial  Theorem 311 

Indeterminate  Coefficients 814 

SECTION  XXXVI. 
Summation  and  Interpolation  op  Series 817 

SECTION  XXXVII. 

Cubic  Equations,  containing  only  the  Third  and  Second  Powers  824 

Cubic  Equations,  containing  only  the  Third  and  First  Powers  329 

Problems  in  Cubic  Equations,  &c 332 

Miscellaneous  Questions 334 

Table  containing  the  Logarithms  of  Numbers  from  1  to  10,000  345 


ALGEBRA 


SECTION    I. 

DEFINITIONS   AND   NOTATIONS. 

Article  1  •     Algebra  is  the  art  of  computing  by  symbols. 

2.  In  Arithmetic  we  represent  quantities  and  perform  cal- 
culations by  figures,  each  of  which  has  a  known  and  definite 
value. 

3.  In  Algebra  we  employ  the  letters  of  the  alphabet,  and 
other  characters,  the  value  of  which  is  either  known  or  un- 
known, according  to  the  conditions  of  the  problems. 

4.  Those  quantities  whose  values  are  given  are  called  known 
quantities ;  and  those  whose  values  are  not  given  are  uTiknown 
quantities. 

5.  The  symbols  used  to  denote  known  quantities  are,  gener- 
ally, the  first  letters  of  the  alphabet  in  the  small  or  Italic 
character,  as  «,  3,  c,  d^  &c. ;  and  those  used  to  denote  unknown 
quantities,  the  last  letters,  as  w,  x^  y,  z. 

6.  In  addition  to  the  above,  which  are  the  more  common 
symbols,  capital  letters  may  be  used,  as  A^  B,  C,  D,  &c.,  or 
letters  of  the  Greek  alphabet,  as  «,  ^,  y,  d,  s,  'c,  &c.  In  ex- 
tensive operations,  the  use  of  these,  or  some  other  suitable 
characters,  is  sometimes  very  convenient. 

7.  Sometimes  quantities  are  expressed  in  Algebra,  as  in 
Arithmetic,  by  figures  instead  of  letters. 

8.  When  a  quantity  is  doubled  or  trebled,  or  multiplied  any 


10  ALGEBRA. 

number  of  times,  the  number  of  multiplications  is  usually  ex- 
pressed by  a  numerical  figure  or  figures.  Thus,  let  a  denote  a 
certain  quantity,  and  la  will  denote  twice  the  same  quantity,  3a 
three  times  the  same  quantity,  &c. 

9i  The  figures  or  letters  prefixed  to  any  symbol,  and  denoting 
the  number  of  times  the  quantity  represented  by  the  symbol  is 
taken,  are  called  the  coefficieiit.  Thus,  in  4a,  lb,  and  4:ax,  the 
coefficients  are  4,  7,  and  4(2. 

10,  A  quantity  which  has  no  figure  prefixed  to  it  is  considered 
as  having  a  unit  for  its  coefficient.     Thus,  a  is  the  same  as  \a. 

1 1 ,  Quantities  represented  by  the  same  symbol  or  letters,  and 
of  the  same  power,  are  called  like  quantities ;  and  those  repre- 
sented by  difierent  symbols  or  letters,  or  by  the  same  letter  of 
different  powers,  unlike  quantities.  Thus,  Za,  4a,  and  ba,  are 
like  quantities ;  and  Sa,  43,  and  5c,  unlike  quantities.  In  like 
manner,  Sab,  Aab,  and  Qab,  are  like  quantities ;  and  dab,  4ac, 
bdc,  and  Qmn,  are  unlike  quantities. 

12,  Besides  the  symbols  and  figures  used  to  denote  quantity, 
there  are  certain  signs,  which  are  used  to  express  the  difi"erent 
relations  between  quantities,  and  the  operations  to  which  these 
quantities  are  subjected.  These  signs  are  the  same  as  are  often 
employed  in  Arithmetic,  but,  in  Algebra,  they  are  indis- 
pensable. 

lej,  The  sign  =  is  that  of  equality,  and  denotes  that  the  two 
quantities  between  which  it  is  placed  are  equal  to  each  other. 
Thus,  a=2b  signifies  that  a  is  equal  to  23. 

14t  The  sign  -}-  is  called  plus,  and  signifies  addition.  Thus, 
a-{-b  signifies  that  a  is  to  be  added  to  b. 

15.  The  sign  —  is  called  minus,  and  signifies  subtraction. 
Thus  a — b  signifies  that  b  is  to  be  subtracted  from  a. 

16.  Sometimes  both  the  signs  -|-  and  —  occur  before  the 
same  quantity,  as  a:^x,  in  which  case  they  signify  that  the 
quantity  may  be  either  added  or  subtracted,  or  that  it  is  doubt- 
ful which  operation  is  to  be  performed. 

17»  The  sign  X  signifies  multiplication.     Thus,  «X^  denotes 


DEFINITIONS    AND    NOTATIONS.  11 

that  a  is  to  be  multiplied  by  h ;  and  ay^by^cy^d,  that  the  quan- 
tities a,  b,  c,  and  d,  are  to  be  multiplied  together.  This  sign  is 
read  into.  Thus,  ay^b  is  to  be  read,  a  into  b.  Sometimes  a 
single  point  is  substituted  for  X*  Thus  a.b  signifies  that  a  is 
multiplied  by  b. 

18,  The  sign  -r-  signifies  division.  Thus,  a-r-b  signifies  that 
a  is  to  be  divided  by  b. 

19,  Division  is  also  represented  by  placing  the  divisor  under 
the  dividend,  in  the  form  of  a  fraction.     Thus,  -  signifies  that  a 

is  to  be  divided  by  b ;  and -,  that  a — b  is  to  be  divided  by 

a-\-b 

a^b. 

20,  The  sign  ]>,  standing  between  two  quantities,  denotes 
that  the  one  before  it  is  greater  than  the  one  after  it.  Thus, 
a'^b  signifies  that  the  quantity  a  is  greater  than  the  quantity  b. 

21,  On  the  other  hand,  the  sign  <[  denotes  that  the  quantity 
before  it  is  less  than  the  one  after  it.  Thus,  b<Ca  signifies  that 
b  is  less  than  a. 

22,  The  sign  .  • .  signifies  therefore.  Thus,  a=b  .  • .  3«=15, 
is  read,  a  is  equal  to  5,  therefore  3(Z  is  equal  to  15. 

23,  The  signs  :  :  :  :  denote  proportion.  Thus,  a  :  b  :  :  c  :  d 
is  to  be  read,  as  a  is  to  3,  so  is  c  to  ^  ;  and  the  signs,  placed  in 
their  order,  indicate  that  a  has  the  same  ratio  to  b  that  c  has 
to  d. 

24,  The  sign  /v/,  called  the  radical  sign,  signifies  the  square 
root  of  the  quantity  which  follows  it ;  or,  that  the  root  of  the 
quantity  is  to  be  extracted.  Thus,  /sj a  denotes  that  the  square 
root  of  a  is  to  be  extracted. 

25,  By  placing  a  figure  above  the  sign,  thus,  /\/,  it  is  made 
the  radical  sign  of  any  root  whatever.  Thus,  /s/a  signifies  the 
cube  root  of  a  ;  ^a,  the  fourth  root  of  a  ;  j^a,  the  fifth  root ; 

Vfi5)  the  wth  root,  &c. 

26,  The  power  of  a  quantity  is  denoted  by  a  figure  placed 


12'  ALGEBRA. 

above  it  at  the  right.     Thus,  c^  signifies  the  second  power  of  a  ; 
a?  the  third  power  of  a  ;  a^  the  fourth  power,  &c. 

27.  In  operating  with  unknown  quantities,  it  is  frequently 
necessary  to  express  the  root  of  a  certain  power  of  a  quantity  ; 
as,  for  instance,  the  4th  root  of  the  3d  power  of  a.  In  this 
case,  a  fraction  is  to  be  used ;  the  numerator  denoting  the 
power  to  which  the  quantity  is  to  be  raised,  and  the  denomi- 

2 

nator  the  root  of  the  power.     Thus,  a^  denotes  the  cube  root  of 

the  second  power  of  a  ;  b'^,  the  fourth  root  of  the  sixth  power 
of  b.     By  inverting  the  fraction,   and  writing   it   before   the 

radical  sign,  we  may  represent  the  same.     Thus,    ^a,    ^b, 

2        6. 

=a^,  b^. 

28.  When  a  quantity  is  represented  by  a  single  letter  or 
numeral,  or  several  letters,  placed  one  after  another  without  the 
sign  _j_  or  —  between  them,  it  is  called  a  simple  quantity. 
Thus,  <2,  be,  cde,  Sab,  are  simple  quantities. 

29.  When  a  number  of  simple  quantities  are  connected  by 
the  signs  +  or  — ,  the  result  is  a  compound  quantity.  Thus, 
a-\-b,  be-{-cd,  4:a-\-5cd — x,  are  compound  quantities. 

30.  A  term  is  a  single  letter  or  numeral,  or  several  letters  or 
numerals,  which  are  not  separated  by  the  sign  +  or  — .  Thus, 
in  the  compound  quantity  a-\-b,  a  and  b  are  the  terms.  So  in 
xy — y-\-z,  xy,  y  and  z,  are  the  terms. 

81.  When  two  or  more  members  of  a  compound  quantity  are 
to  be  subjected  to  the  same  operation,  in  which  they  are  to  be 
regarded  as  one  whole,  they  are  connected  by  a  line  drawn  over 
them,  called  a  vinculum,  or  by  enclosing  them  in  a  parenthesis. 
Thus,  when  we  are  to  multiply  a-\-b-\-c  by  any  number,  as  3, 
we  write  a-{-b-\-cXS,  or  (ft-l-^+c)x3,  or,  more  simply,  3(a+ 

h-\-c).  So  x^yXfV^,  01"  {^-\-y)  (2/+^)  signifies  that  x-{-y  is 
to  be  regarded  as  a  whole,  and  multiplied  by  y-\--^  i^^^Vi.  also  as 
a  whole  ;  whereas,  if  the  line  or  parenthesis  were  not  employed, 
a^bX^  would  denote  that  b  only  is  to  be  multiplied  by  3,  and 
the  result  would  be  a-\Sb. 


DEFINITIONS  AND  NOTATIONS.  13 

32.  When  two  or  more  quantities  are  multiplied  togetlier, 
each  quantity  is  called  a  factor.  Thus  in  ah,  a  and  h  are  called 
factors ;  so,  in  cde,  c,  d  and  e,  are  severally  called  factors. 

33 1  A  composite  number  is  one  which  is  produced  by  the 
multiplication  of  two  or  more  quantities  or  factors  into  each 
other.  Thus,  the  quantity  abc  is  a  composite  one,  the  factors 
of  which  are  a,  b,  and  c. 

34.  Quantities,  which  have  the  sign  -{-  before  them,  either 
expressed  or  implied,  are  called  positive  or  affirmative  quantities. 
Thus  -\-a,  -\-b,  or  a,  b,  are  positive  or  affirmative,  the  sign  -j- 
being  always  implied  before  a  quantity  which  has  no  express 
sign  prefixed. 

35.  Quantities,  which  have  the  sign  —  prefixed  are  called 
negative  quantities.  Thus,  — a,  — b,  — 3,  are  negative  quan- 
tities. 

36.  Where  the  signs  are  all  positive  or  all  negative,  they  are 
called  like  signs. 

37.  When  some  of  the  signs  are  positive  and  others  negative, 
they  are  said  to  be  urdike. 

38.  WTien  a  quantity  consists  of  one  term  it  is  called  a 
monomial,  as  a,  ab,  Sxy,  being  the  same  as  a  simple  quantity. 

39.  When  a  quantity  consists  of  two  terms  it  is  called  a 
biTwmial.  Thus  a-{-b,  x-\-y,  are  called  binomial  quantities,  and 
a — 3  a  residual  binomial. 

40.  When  a  quantity  consists  of  three  terms  it  is  called  a 
trinomial.     Thus,  a-\-b-\-c,  and  x-\-y-\-z,  are  trinomials. 

41 .  When  a  quantity  consists  of  any  number  of  terms  greater 
than  three  it  is  called  a  polynomial.  Thus,  a-\-b~\-c-\-d,  and 
vj-\~x-\-y-\-z,  are  polynomials. 

42.  The  power  of  a  quantity  is  its  square,  cube,  biquadrate, 
&c.,  called,  also,  its  second,  third,  fourth  power,  &c. ;  as  c^,  a^, 
a*,  &c. 

43.  The  index  or  exponent  of  a  quantity  is  the  number  which 
denotes  its  power  or  root. 

2 


14  ALGEBRA. 

Thus,  — 1  is  the  index  or  exponent  of  a~^ ;  2  is  the  index 

-|  1 

of  a^;  J-,  of  <z^,  or  yv/a;  and  m  and  -,  of  a""  and  a". 

n 

44.  When  a  quantity  appears  without  any  index  or  exponent, 
it  is  always  understood  to  have  for  it  unity,  or  1. 

Thus,  a  is  the  same  as  <z^,  2x  is  the  same  as  2x^ ;  the  1  in 
such  cases  being  usually  omitted. 

45.  A  rational  quantity  is  that  which  can  be  expressed  in 
finite  terms,  or  without  any  radical  sign  or  fractional  index ;  as 

a,  or  -75-,  or  bc^,  &c. 
o 

46.  An  irrational  quantity  is  that  which  has  no  exact 
root,  or  which  can  only  be  expressed  by  means  of  the 
radical  sign,  or  a  fractional  index;  as  /s/a,  or  22,  /^/~^,  or 
fl^,  &c. 

47.  A  square  or  cube  number,  &c.,  is  that  which  has  an 
exact  square  or  cube  root,  &c. 

9  8 
Thus,  4  and  —^c^  are  square  numbers ;  and  64  and  -— a^  are 

10  ^i 

cube  numbers,  &c. 

48.  A  measure  or  divisor  of  any  quantity  is  that  which  is 
contained  in  it  some  exact  number  of  times,  without  a  re- 
mainder. 

Thus,  3  is  the  measure  or  divisor  of  6,  la  is  a  measure  of 
35a,  and9«^of27«-^^. 

49.  Commensurable  numbers  or  quantities  are  such  as  have  a 
common  measure  or  divisor,  or  that  can  be  each  divided  by  the 
same  quantity  without  a  remainder. 

Thus,  6  and  8,  2/^/2  and  3^2,  ba%  and  la%  are  com- 
mensurable quantities  ;  the  common  divisors  being  2,  a/2,  and 
a^b. 

50.  A  priine  number  is  that  which  has  no  exact  divisor, 
except  itself  and  unity;  as  1,  2,  3,  5,  7,  11,  13,  17,  &c. ;  and 
the  intervening  numbers,  4,  6,  8,  9,  10,  12,  14,  and  16,  are  com- 
posite 72umbers. 


DEFINITIONS  AND  NOTATIONS.  15 

51*  Two  or  more  numbers  are  said  to  be  incommensurable^  or 
'prime  to  each  other,  when  thej  have  no  common  divisor  except 
unity ;  as,  2  and  3,  5  and  7,  17  and  19,  &c. 

52.  One  quantity  is  called  the  multiple  of  another,  when  the 
former  contains  the  latter  a  certain  number  of  times  without  a 
remainder. 

Thus,  15a  is  a  multiple  of  ba,  and  Qa  of  oa. 

53.  The  reciprocal  of  any  quantity  is  unity  divided  by  that 
quantity. 

The  reciprocal  of  any  fraction  is  that  fraction  inverted. 

Thus,  the  reciprocal  of  c  or  -  is  - ;  the  reciprocal  of  -  is 

I       a  b 

-,  and  the  reciprocal  of  —^  is  — r-r. 
a  a — 0      a-\-b 

54.  A  series  is  a  rank  or  succession  of  quantities,  which 
usually  proceed  according  to  some  certain  law ;  as  \-{-\a-{-^dr 

PRACTICAL   EXAMPLES. 

55 1  In  calculating  the  numerical  values  of  the  following 
Algebraic  Expressions,  let  a=6,  ^=5,  c=4,  c?=l,  and  e=0. 

1.  «'24-2«5—c+^=36+60— 4+1=93. 

2.  2a^—3a^i+c^=432— 540+64=— 44. 

3.  a-X(a+^)—2a5c  =  36xll— 240  =  156. 

4.  2aV^^— ac+V2«c+?=12Xl+8=20. 

5.  Za^/2ac-\-c\  or  3«(2ac+c')2-=18V64=144. 

6.  V(2a'-/v/6ac+e^)=V(72-V144)=V60. 

7.  V(«'^-f4^'—5Vc^)  =  V (180+100-10)= V270. 

8.  V(«^'+233— 5V^c)=V(150+250— 10)  =  V390. 

9.  |+f+^f^=?i+JL?!+f =4+10=14. 
6^+4e     V2«c+c'^      6   '  a/64      6   '    8  ' 

2^3— 5c      /3a— J^>cU_40        [18— 10_40        I4_ 
•  4^»-10^"^  V5a-12(^/  ~'10"^aJ30-12~~I0"^aJ9~ 
4|. 
11.  2a'+3^c-5=127. 


16  ALGEBRA. 

12.  ba''b—10ab'-\-27e=z—Q00. 

13.  7a''-{-{b-c)X{d—e)=2bS, 

^  .    alP'      ,     a — h  ,  ,-„  „ 

14.  — Xd ; — \-27a^e=^Q^. 

c  a 

15.  S^c-\-2a^{2a-^b-d)=M. 

16.  a/s/{a'-\-e)-\-Sbc{a'—b'')=mQ. 

17.  Sa^b-j-W  {c''—A/2ac-{-c')—oe=D42. 


SECTION   II. 

ADDITION. 


Art.  5^»  Addition  in  Algebra  is  the  connecting  together  of 
several  quantities  by  their  appropriate  signs. 

57.  The  operation  consists  in  collecting  into  one  term  all  the 
like  quantities,  and  so  arranging  the  several  terms,  thus  obtained, 
as  by  signs  to  indicate  the  proper  sum  of  all  the  quantities,  both 
like  and  unlike. 

58.  Addition  in  Algebra  embraces  three  cases. 

I.  When  the  quantities  are  alike,  and  their  signs  alike  also ; 
as,  a,  Sa  ;  or,  — b,  — 43. 

II.  When  the  quantities  are  alike,  and  their  signs  unlike ;  as, 
33,  -53. 

III.  When  the  quantities  are  unlike,  some  having  like  and 
others  unlike  signs  ;  as,  Sa,  43,  — ix. 

Case  I. 

59.  When  the  quantities  are  alike,  and  their  signs  alike. 

Rule.  Add  together  the  coefficients  belonging  to  the  like 
quantities,   and   place    their    sum    before    the    common    letter 


ADDITION.  17 

or  letters^  with  the  common  sign  prefixed  ;  and  the  result  will  be 
the  sum  required. 

Thus,  let  it  be  required  to  add  together  3a3,  Idby  8^3,  the 
operation  will  be  as  follows :  — 

Zab 

'Jab 

Sab 


ISab. 


The  reason  of  this  rule  is  obvious ;  for,  since  ah,  whatever  be 
its  value,  must  represent  the  same  quantity  in  every  instance, 
it  is  evident  that  3  times,  7  times,  and  8  times  the  same  quan- 
tity, will  make  18  times  the  same. 

In  like  manner,  let  it  be  required  to  add  together  — 7b,  — 53, 
and  — 63. 

-73 
-53 
—63 


—183. 

EXAMPLES. 

(1) 

(2) 

(3) 

(4) 

(5) 

3a 

7A 

— Sax 

4:xy' 

Za-  If 

4a 

bk 

—4:ax 

xf 

4a-  Sif 

Qa 

Sh 

—  ax 

Zxf 

Qa—     y" 

a 

Sh 

— 2ax 

Ixf 

a —  6?/^ 

5a 

h 
24A      - 

—lax 

2xf 
Vlxf 

5a—  2y^ 

19a 

-VJax 

19a-142/2 

(6) 

(7) 

(8) 

(9) 

(10) 

7x 

14a3c 

^y 

—  \mn 

bh-\-  x 

4^ 

lla3c 

y 

—  3??m 

h-\-2x 

11a; 

babe 

y 

—     mn 

2h-i-4.x 

9x 

4a3c 

oy 

— Yhnn 

h-i-  X 

X 

abc 

4y 

—    mn 

Ih-^-Qx 

18  ALGEBllA. 

11.  Add  7a,  11<2,  <2,  4a,  Qa,  and  3a  together.         Ans.  32a. 

12.  Add  4Ji,  Qk,  k,  k,  llh,  and  7k  together.  Ans.  dOk. 

13.  Add  together  {Sa'—b),  (7a'— 43),  and  (a'—b). 

A?is.  lla^— 63. 

14.  \Yhat   is   the   sum    of  3/v/a-,   4/\/a-,  a/o^,  7a/ a*^,  and 
2a/ a'?  ^^w.  17/v/«'- 

15.  Add  together  3/s/ci-{-b,  Qa/o^S,  hj a-\-b^  and  12/\/a+3. 


^?Z5.  '22/s/a-{-b. 

Case  II. 

60.  When  the  quantities  are  alike,  and  have  unlike  signs. 

Rule.  Add  all  the  affirmative  coefficients  into  one  sum,  and 
those  that  are  negative  into  another  ;  then  subtract  the  less  of 
these  results  from  the  greater,  and  prefix  the  sign  of  the  greater 
to  the  difference,  annexing  the  common  letter  or  letters. 

Required  the  sum  of -j-7a:r,  — Aax,  — 3aa;,  -\-Ylax,  — ax,  and 
-\-ax.  7ax 

—  4:ax 

—  Sax 
17  ax 

—  ax 
ax 


VJax. 


We  find  the  sum  of  the  plus  quantities  to  be  25a:r,  and  the 
sum  of  the  negative  quantities  — ^ax ;  and  the  difference  be- 
tween those  coefficients  is  17,  which  we  prefix  to  ax;  thus, 
17a:;::. 

The  reason  of  this  Rule  is  obvious,  when  we  consider  that 
two  equal  quantities,  the  one  with  a  positive  and  the  other  with 
a  negative  sign,  exactly  cancel  each  other,  so  that  their  sum  is 
nothing.  Of  course,  then,  when  two  like  quantities  of  opposite 
signs  are  not  equal,  the  difference  between  them  must  be  the 
proper  sum,  which  will  be  positive  or  negative  according  to  the 
affirmative  or  negative  character  of  the  larger  quantity. 


ADDITION. 

1 

EXAMPLES, 

(1) 

(2) 

(3) 

(4) 

(5) 

7a 

—  G»^ 

Aax 

ISn 

7  a—     mp^ 

—  Sa 

771 

— Sax 

n 

a-{-  Gmp'^ 

a 

-11m 

ax 

— 20?z 

— 11a —  S/Tzp^ 

—  5a 

bm 

—7  ax 

6?i 

8a+ll777/ 

11a 

—     m 

ax 

8  71 

—   da —    7777j9'^ 

a 

20m 

12ax 

—       71 

18a — lb77ip' 

12a 

S?7l 

Sax 

771 

14a —  977ip' 

(6) 

(7) 

(8) 

(9) 

(10) 

6y 

477171 

—Sxy 

-  Apn 

Sxy —     77iy 

-  72/ 

7n7l 

7xy 

pn 

3z7/-j-     ?7iy 

4:y 

S77271 

—4:xy 

p7i 

—  11a;  7/ — 18777^^ 

-112/ 

1877271 

—  ^y 

-llp7i 

—  4iXy-\-  977vp 

9y 

777171 

9xy 

7pn 

—     8Z7/ —    3777^^ 

—  ^y 

Sm7i 

-Sxy 

p7l 

12a:7/-}-12777> 

19 


11.  Add   4+a^a;,    Q—a^x,  3+6a':i',    15— 5a^:7;,  3+a^:c,  and 
e+Ta'^a:  together.  Ans.  37+9a'a:. 

12.  Add    14aa;— 6?/,    7ax-]-y,     bax — 7y,    9ax — lly,    and 
Sax-\Sy  together.  A7is.  ASax — 20y. 

13.  Add  3a— 43+6c,  7a+113— 3c,  8a4-^— 7c,  and  a— 113 
+15c  together.  Atis.  19a— 33-j-llc. 

14.  Add   16^;'^— 5?/— 16,  Sx^+^y^-b,  x^'-^Sf-ST,  x'—y^ 
+7,  Qx''-{-77/—ll,  and  2x''—Sy'—21  together. 

Atis.  29a:--f5?/^— 83. 

15.  Add  5a— 3,  33-|-8c,  4a— be,    ba—bb—c,    7a— Qc,  and 
lla+43— 7c  together.  Ans.  32a4-3— 16c. 

Case  HI. 

61.  "When  the  quantities  are  unlike,  some  having  like  and 

others  unlike  signs. 

It  is  evident  that  unlike  quantities  cannot  be  united  into  one  ; 
or  otherwise  added  than  by  means  of  their  signs. 


J 


20 


ALGEBEA. 


Thus,  for  example,  if  a  be  supposed  to  represent  20,  and  h 
12,  then  the  sum  of  a  and  h  can  be  neither  twice  20  nor  twice 
12,  but  it  must  be  20+12=32,  that  is,  a-^h. 

62i  Hence,  to  add  unlike  quantities,  we  have  the  following 

KuLE.  Collect  all  the  like  quantities  together^  as  in  Case  II., 
and  write  down  those  that  are  unlike,  one  after  another,  with 
their  proper  signs. 

63i  When  several  quantities  are  to  be  added  together,  it  is 
immaterial  in  what  order  they  are  written. 

Thus,  a-{-b — c,  a — c-\-b,  — c-\-a-\-b,  are  equivalent  expres- 
sions. 


(1)  ^ 

^JlXAMPLES. 

(2) 

Zax 

7  a-\-7m 

A^mn 

-L 

6  a — 7x 

-W 

Axy — bm 

Ixy 

nn — Qy'^-{-7xy. 

^x  +82:?/ 

2>ax-\-A7s 

lU-\-12xy+2m—Ax, 

(3) 

(4) 

/           (5) 

9^V 

14:ax  —  2x'^ 

4fl^2:— 130+3a:^ 

—"Ix'^y 

^ax"^-}-  Sxy 

5^2  +3aa:+9ar2 

Saxy 

Sy^  —  Aax 

7xy—Ax\^^' 

-4.xf 

dx''  +26 

^x—AO  —^x" 

6.  Add  together  a-\-x  and  y — c.  Ans.  a — c-\-x-\-y. 

7.  Add  3a+3— 10,  c—d—a,  _4c+2£z— 35— 7,  and  Ax'-^-b 
—18771  together.  An^.  4^^—23— 12— 3c— ^+4:i;'^—18??i. 

8.  Add  7a— by^,  S/s/x-\-2a,  by^ — a/x,  and  — 9a-{-7/s/x  to- 
gether. Ans.  14:a/x. 

9.  Add   4m7i-\-Sab — 4c,    Sx — 4:ab-\-2mn,   and  o?n' — 4p   to- 
gether. Ans.  6  mn — ab — 4c+3a:+3m'^ — 4:p. 

10.  Find  the  sum  of  Sa^-\-2ab-\-4:b\  ba''—Sab-\-b'',  —a''-\-bab 
—b-,  lSa^—20ab—ldP,  and  Ua''—Sab-{-20b-. 

Ans.  39a''—2'iab+bb\ 
a  -^ 

/ 


SUBTRACTION.  21 

11.  Find  the  sum  of  \.x^—^c^—hax^-\-^d^x,  ^d^-\-%x^-\-^^cix^ 
^1a^x-Vlx^-\-V^a£--\^d}x,  \Zax'—'nd}x-\-\U\  %d}x—1^a' 
-\-Vlx\  and  ^\(i~x—1x^—Z\ax~—lx\  Ans.—7x^—a\ 

64.  Coefficients,  whether  figures  or  letters,  that  are  common 

to  several   terms,  may  be  connected   with   them   by  a   paren- 
thesis. 

(12)  (13) 

-   Add  amx-{-2dy  hy-\-4cmx 

^cx — ^dy  my-\-  Mx 

Zax-\-by  'hiy — ^mx 


{am-\-'lc-{-^a)x-\-{b—d)y.  {Ji-\-'m-\-^n)y-\-{M—bm)x. 

14.  Add  4:ax — b7ny,  odx-\-77iy,  and  7mx-\-4?ny,  together. 

A71S.     {4ia-{-dd-\-77n)x-{-{7n — 7n)y. 

15.  Add  ohz — t)x,  A.mz-\-7ix,  and  bai — ^px,  together. 

Atis.     {^h-\-^m-\-ha)z-\-{7i—b—4:p)x. 

to  -/5-  -  r 

SECTION  III.  a  o    ~    ^      '      ^  * 

SUBTRACTION. 

Art.  65  0  Subtraction  is  the  taking  of  one  quantity  from 
another,  or  the  method  of  finding  the  difierence  between  any  two 
quantities  or  sets  of  quantities  of  the  same  kind. 

66.  If  it  be  required  to  subtract  10 — 7  from  12,  we  might 
first  subtract  7  from  10=3,  and  take  the  3  from  12=9 ;  or 
we  might  take  the  10  from  12,  and  the  remainder  2  must  neces- 
sarily be  increased  by  7  to  produce  the  correct  result. 

If  from  a  we  wish  to  subtract  c — d,  we  first  subtract  c,  and  it 
gives  a — c.  This  quantity,  since  we  have  taken  d  too  much 
from  a,  is  too  small  by  d.  Therefore  d  must  be  added,  thus, 
a — c-}-d. 


22  ALGEBRA. 

67.  If  a  simple  quantity  is  to  be  taken  from  another  simple 
quantity,  it  is  only  necessary  to  write  them  one  after  the  other ; 
thus,  if  8  is  to  be  taken  from  15,  it  may  be  expressed  thus, 
15_8=7. 

If  it  were  required  to  subtract  h  from  c,  it  should  be  written 
thus,  a — h  ;  but  if  we  were  to  subtract  a — h  from  cA^d^  it  is 
evident  that  if  only  a  were  to  be  taken,  it  would  be  written 
thus,  c-\-d — a.  But  this  evidently  gives  a  result  too  small ;  for 
a  was  to  be  lessened  by  h  before  the  subtraction.  Therefore,  as 
the  remainder  is  too  small  by  5,  we  must  add  l  to  the  remainder, 
which  will  give  c-f-^ — a-^l ;  for  it  makes  no  difference  in  the 
result  whether  the  minuend  be  increased  or  the  subtrahend 
lessened. 

Subtract  7—4  from  13.  Taking  7  from  13  leaves  6 ;  but  6 
is  too  small,  for  the  7  should  have  been  lessened  by  4,  and  we 
must  either  subtract  the  4  from  the  7  before  the  operation,  or 
add  it  to  the  remainder;  and,  if  added  to  the  remainder,  the 
expression  will  be  thus,  6-}-4=10. 

68.  We  therefore  see  the  propriety  of  the  following 

Rule.  Change  the  signs  of  all  the  quantities  to  be  subtractedj 
and  proceed  as  in  Addition. 

SIMPLE   QUANTITIES. 

(1)  (2)  (3)  (4)  (5)  (6) 

From  +7a         -16^         +17^         —2%        -j-153        —6c 
Take  +2a         —  bx         +8^         -18^         +  ^^        —  c 

4-5a         —11a;         +  9d         —llg        +  85        —5c 

The  above  questions  are  performed  as  in  Arithmetic,  the 
minuend  being  the  larger  number,  and  having  the  same  sign  as 
the  subtrahend. 

(7)         (8)  (9)         (10)       (11)         (12)        (13) 

From  —  Sa      -\-  7x      +18?/      —35      —  7c      +8(7     —(Jh 
Take   —15a      +14a;      -\-20y      —7b      —15c      +11(7      —Sh 

+  7a     —  7x      —  2y      +45      +  8c     -  3(7      -\-2h 


J 


OrJ  i 


SUBTRACTION.    *  23 

In  these  examples  the  minuend  is  taken  from  the  subtrahend, 
and  all  the  signs  in  the  subtrahend  are  changed. 

(14)        (15)      (16)       (17)        (18)         (19)         (20) 
From  +27«     —  63     —7c      +%      ~r'^'^^^      —  bx      —  7y 
Take   —13a     +185     —  c      —9g      -Ibk      +17a;      —Iby 

+40a     —243     —Qc        17g      +2Qh      —22^     +  87/ 

In  these  examples  we  change,  mentally,  all  the  signs  in  the 
subtrahend,  and  then  proceed  as  in  addition.  These  questions 
may  all  be  proved,  as  in  Arithmetic,  by  adding  the  remainder  to 
the  subtrahend. 

COMPOUND   QUANTITIES. 

69.  The  same  rule  must  be  observed  in  subtracting  compound 
quantities  as  in  simple  quantities ;  that  is,  all  the  signs  of  the 
quantities  to  be  subtracted  must  be  changed,  the  signs  +  to  — , 
and  the  signs  —  to  -}- ;  we  then  proceed  as  in  addition. 

(1)  OPERATION. 

From    ab-{-  cd —  ax —  7  ab-\-  cd —  ax —  7 

Take  4:ab—Scd-{-4:ax—lb  — 4:ab-\-ocd—4:ax-\-lb 

—Sab-{-4:cd—bax-\-  8 

70.  It  is  a  better  way  for  the  pupil  to  conceive  the  signs  in 
the  subtrahend  changed,  but  to  let  them  remain  without  alter- 
ation, otherwise  it  might  be  difficult  to  correct  errors  that  might 
arise  in  the  operation. 

(2)  (3) 

From  7x+5?/— 3a—  6A  1abc—llx-\-by—  48 

Take     x—ly-]-ba-\-llh  llabc-\-  3a;-}-7?/-j-100 

Qx^l2y—^a—VJk  _4a3c— 142-— 2?/— 148 

(4)  (5) 

From    14^—42+  9?/     -{-x  ^x—babc—Qh—b\ 

Take  —U~lz-\-41y—Vlx  19a;— 7a3c— 8A-|-  1 

17y^_j_32_32?/-j-18z  _10:r-j-2a3c+2/^— 52 


24  •         ALGEBRA. 

(6)  (7) 

From    "dxy—a^  — SA^—     y'  7x''—a'b'-{-  ly'-^-W 

Take  —xij—a'  ^-7/^'--10^/^  x'-^a?b'—\ly'—  h' 

(8)  (9) 

From  b^ax^~-2>f  -\-la^—l  Sor^-f  y'-\-  V7A+  5 

Take  3V«^'+  y"  — 5dz^+7  4:cL3y5_  ^7^_  6 

2V«^'— 4r+12aL8  4a;^+4y^+2V  7^+11 

10.  From  3«— 55+6/^— ^  take  a-]-b—ld. 

Am.  2a—Qb-\-m-]-M. 

11.  From  Zlx'—Zy''-^ab  take  IT^'^+S?/^— 4a3-]-7. 

J[?z5.  14jc^— 82/^4-5a^— 7. 

12.  From  5/+14i— 9^^  take  _3/+73— 15^. 

Am.  8/-|-73-|-6^. 

13.  From  lla—lb^c  take  ^+7^— 3c+ll. 

^?z^.  10a— 143+4c— 11. 

14.  From  77i^-\-3n^  take  — 47?^- — 67^^-["71''^• 

^715.  5^2+97^3— 71.r. 

15.  From  31a— 15a;— 7  take  2a—2bx^if. 

Am.  29a-\-l0x—f—7. 

16.  From  abc^ — xy^  take  — Qabc^-^Sxy^ — 7k. 

Am.  labc^—^xf-\-7h. 

17.  From  llcA'— 5  take  5c/i'— 5+47^:. 

Ans.  6cA- — 47a;. 

18.  From  m7i?-\-j£t  take  — lm7i?-\-4:^x—y''\ 

Am.  8m?i'+Z;?— 48a;+2/\ 

19.  From  47a^/z— 37+962^^  take  7abh. 

Am.  40a3A— 374-96?/". 

20.  Take  7^;^+ ^771  from  8a;y+17. 

Ans.  xry^ — hm-\-\l. 

21.  Take  5^^— 3c+597?i  from  1131 

Am.  6^2.^  3c— 59771. 

22.  Take  6a— 33— 5c  from  6a+33— 5c+l. 

Am.  63+1. 


SUBTRACTION.  25 

23.  Take  4:lx''-{-7f+abc  from  m\ 

Ans.  7)-^ — 41a;' — l^f — ahc. 

24.  Take  x^  from  —Vlx^-\-\^y—a^h. 

Am.  --\W^\^y—a-\-h. 

25.  Take  a—h  from  a-^h.  Ans.  -f  1h. 

26.  From  %xz  take  a:z— 7/i— 57?^^-f-7. 

Ans.  8:1:2+7^4- 5??z3— 7. 

27.  From  Il/i7?i-|-8?z2  take  x^—f. 

Ans.  Wlim-\-^7i} — :>?-\-if' 

28.  From  a-\-h  take  a — 3,  and  gj — 3,  and  — a4-^- 

Ans.  +23. 

29.  From  gj — h — c  take  — a-\-h-\-c  and  fz — l-^-c. 

Ans.  a — h — 3c. 

71.  When  similar  quantities  that  are  to  be  subtracted  have 
literal  coefficients,  the  operation  may  be  performed  by  placing 
the  coefficients  with  their  proper  signs  within  a  parenthesis,  and 
then  subjoining  the  common  quantity ;  thus. 

From  ay—h  From         ax^-\-gy'^ 

Take  dy — c  Take  bx^ — hy"' 

{a-d)y+c-h  {a-b)x'-}-{g+h)f. 

72,  If  a  set  of  quantities  enclosed  in  a  parenthesis  is  com- 
bined with  others  by  means  of  the  sign  + ,  the  parenthesis  can 
have  no  effect  upon  the  result,  and  may,  therefore,  be  retained 
or  not,  at  pleasure. 

Thus,  a-{-{b-\-c)  is  evidently  equivalent  to  a-\-b-\-c ;  for  it 
can  make  no  difference  whether  b  and  c  be  first  added  together, 
and  their  sum  then  be  added  to  a,  or  the  sum  of  the  three  quan- 
tities, a,  b,  c,  be  taken  at  once. 

Again,  x — y-\-{b — z)  will  amount  to  the  same  thing  as  x — y 
-\-b — z;  for  it  is  immaterial  whether  b — z  he  added  to  x — y 
at  once,  or  b  be  added  to  it  first,  and  from  the  result  z  be 
subtracted. 

The  subtraction  of  a  polynomial  may  be  indicated  without 
performing  the  operation,  by  inclosing  the  quantity  to  be  sub- 
tracted in  a  parenthesis,  and  prefixing  the  sign  — . 


26  ALGEBRA. 

If  we  wisli  to  subtract  7a—^x-\-Qy  from  11a — 2x-{-Si/,  it 
may  be  indicated  thus  (ll<z — 2z+8?/)  —  (Ja—bx-j-Qy). 

And  7a — Sb-{-c-\-g — p,  taken  from  10a,  leaves  10a — (la — ob 
\-c-\-g — p) ;  being  equivalent  to  3a-}- 33 — c — g-\-p. 

If,  therefore,  a  quantity  enclosed  in  a  parenthesis  be  com- 
Dined  with  another  by  means  of  the  sign  — ,  the  rule  laid  down 
in  Art.  68  shows  that  the  signs  of  the  terms  of  this  quantity 
must  be  changed  whenever  the  parenthesis  is  removed. 

Thus,  a — (3-f-c)  is  equivalent  to  a — b — c;  because  it  can 
be  of  no  importance  whether  b  be  first  subtracted  from  a,  and  c 
then  be  taken  from  the  remainder,  or  the  sum  of  b  and  c  be 
subtracted  from  a  at  once. 

Consequently,  a  parenthesis,  with  a  negative  sign  preceding  it, 
may  be  introduced  into  any  compound  algebraical  expression, 
provided  the  signs  of  all  the  symbols  comprised  in  it  be 
changed. 

Thus,  a — X — b-{-y  is  equivalent  to  a—x — {b — y),  or  a — {x-\- 
b — y),  or  a-{-y — {b-[-x),  or  y — (x-{-b — a). 

Similar  considerations  will  enable  us  to  dispense  with  the  use 
of  parentheses,  without  altering  the  values  of  the  expressions  in 
which  they  are  found,  when  one  or  more  such  parentheses  are 
included  within  another. 

Thus,  a — [b — {c-^d)]  is  manifestly  equivalent  to  a — [b — c 
— d],  which  is  also  equivalent  to  a — b-{-c-\-d. 

Also,  a — \a-\-b — [a-}-b — c — {a — b-{-c)]\=:a — \a-{-b — [a-\-b 
^c—a^b—c]  \=a—\a-{-b—[2b—2c]  \=a—\a-{-b—2b-\-2c\ 
z=a—\a—b-^2c\=a—a-\-b—2cz=b—2c. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  value  of  the  expression  (1 — 2x-\-Sx'^)-}-{S-\- 
2z— a;-)?  Ans.  4+2:cl 

2.  Reduce  to  its  simple  form  the  expression  5a — 4:b-\-dc-\- 
(^^^a-\-2b—c).  Ans.  2a—2b-{-2c. 

3.  What  is  the  value  of  the  expression  (a — b — c)-\-{b-\-c — d) 
Jr{d-e+f)-^{e-f-g)  ?  Ans.  a-g. 


SUBTRACTION.  27 

4.  Exhibit  a — {h—c)-\-h—{a—c)-\-c — {a — h)  in  its  simplest 
form.  Ans.  — a-{-h-\-oc, 

5.  From  ^x'^f)  take  l{Qc'J^'lxy^y')—{^2xy—x^—f)\ 

Ans.  x^-Y'if'. 

6.  From  Qx'+2y'—{^x'^y'')  take  2x''+^''—{'ix'—if). 

Ans.  t)x^ — 42/^. 

73 1  Algebra  differs  from  Arithmetic  in  the  use  of  negative 
quantities.  In  Algebra,  every  quantity  is  either  positive  or 
negative,  according  as  it  is  affected  with  the  sign  plus  or  minus ; 
and,  as  we  have  observed  above,  whenever  a  quantity  has  not 
either  of  these  signs  prefixed,  the  sign  +  is  understood,  and 
the  quantity  is  said  to  be  positive.  Thus  5,  or  +5,  is  positive; 
but  — 5  is  negative.  Positive  quantities  are  also  called  affirm- 
atives. Some  mathematicians,  in  treating  this  subject,  have 
involved  it  in  much  perplexity,  and,  in  our  opinion,  in  absurd- 
ities, by  considering  — 5,  or  — a,  as  quantities  Ze55  than  nothing  ; 
much  to  the  injury,  if  not  to  the  disgrace,  of  the  science.  But 
the  student  is  to  observe  that  — 5  denotes  just  the  same  number 
and  quantity  as  -\-b,  but  with  the  additional  considerations, 
that  the  former  is  to  be  subtracted^  while  the  latter  is  to  be 
added. 

The  simplest  illustration  of  positive  and  negative  quantities 
may  be  derived  from  a  merchant's  credits  and  debts.  Five 
dollars  is  the  same  sum,  whether  it  be  due  to  him,  or  he  owe 
it  to  another ;  but,  in  one  case  it  may  be  considered  as  'positive 
$5,  for  it  is  an  addition  to  his  property ;  and  in  the  other  as 
negative  $5,  for  it  is  subtracted  from  his  property.  And,  if 
the  sum  of  his  debts  exceeds  the  sum  of  his  credits  by  $1000, 
the  state  of  his  affairs  may  be  represented  by  — $1000 ;  and, 
undoubtedly,  he  is  worse  than  if  he  had  nothing,  and  owed 
nothing.  In  such  a  case,  indeed,  the  man  is  often  said,  in  mer- 
cantile language^  to  be  minus  one  thousand  dollars.  Whereas, 
if  the  sum  of  his  credits  exceeds  the  sum  of  his  debts  by  SIOOO, 
the  state  of  his  affairs  may  justly  be  represented  by  +81000. 
These  opposite  signs,  then,  without  at  all  affecting  the  absolute 


28  ALGEBRA. 

magnitude  of  the  quantities  to  whick  they  are  prefixed,  intimate 
the  additional  consideration  that  those  quantities  are  in  contrary 
circumstances. 


SECTION    IV. 

MULTIPLICATION. 

Art.  74 »  Multiplication  is  the  repeating  of  a  quantity  as 
many  times  as  there  are  units  in  another;  it  is  virtually  the 
same  in  Algebra  as  in  Arithmetic. 

75.  The  multiplicand  and  multiplier  may  be  considered  as 
factors;  and,  in  all  operations,  either  may  be  taken  for  the 
other. 

Thus,  if  6  be  multiplied  by  7,  or  a  by  3,  the  result  is  tho 
same  as  if  7  be  multiplied  by  6,  or  h  by  a. 

76.  When  several  letters  are  written  after  one  another,  it 
implies  that  they  are  all  multiplied  together. 

Thus,  abed  is  the  same  as  ay^by^cy^d ;  and  it  is  immaterial 
in  what  order  they  stand ;  for  abed,  cdab,  and  bdca,  are  synony- 
mous terms. 

77.  Multiplication  is  commonly  divided  into  three  cases. 

I.  When  the  multiplicand  and  multiplier  are  simple  quan- 
tities. 

II.  When  the  multiplicand  is  a  compound  quantity,  and  the 
multiplier  is  a  simple  one. 

III.  When  both  the  multiplicand  and  multiplier  are  com- 
pound quantities. 

Case  I. 

78.  When  the  multiplicand  and  multiplier  arc  simple  quan- 
tities. 


MULTIPLICATION.  29 

Rule.  Multiply  the  coefficients  of  both  terms  together,  and 
to  the  product  annex  the  letters  in  both  factors,  remembering 
that  the  product  of  like  signs  is  plus,  and  of  unlike  signs  is 
minus.  That  is,  plus  (-f-)  multiplied  by  plus  (-}-),  and  minus 
(— )  multiplied  by  minus  ( — ),  give  plus  (-j-)  /  ^^^^  plus  (-}-) 
multiplied  by  minus  ( — ),  and  minus  (  — )  multiplied  by  plus 
(•-|-),  give  minus  ( — ). 

ILLUSTRATION. 

79.  1.  If  a  plus  quantity  is  multiplied  by  a  plus  quantity, 
the  result  will  be  a  plus  quantity.     Thus, 

If  -\-a  is  multiplied  by  -\-b,  it  is  evident  that  -\-a  is  to  be 
repeated  as  many  times  as  there  are  units  in  -{-b ;  that  is,  b 
times  a=.-\-ab. 

2.  If  a  minus  quantity  is  multiplied  by  a  plus  quantity^  or  a 
plus  quantity  by  a  minus  quantity,  the  result  will  be  a  minus 
quantity.     Thus, 

If  — c  is  to  be  multiplied  by  -{-d,  it  is  evident  that  — c  must 
be  repeated  as  many  times  as  there  are  units  in  d ;  that  is, 
d  times  — c= — cd.  The  result  will  be  the  same  if  -|-c  is  mul- 
tiplied by  — d. 

3.  If  a  minus  quantity  be  multiplied  by  a  minus  quantity, 
the  result  will  be  a  plus  quantity. 

To  illustrate  this,  let  a — b  be  multiplied  by  c — d. 

The  product  of  a — b  by  c  is  ac — be ;  but  it  is  evident  that 
this  product  is  as  many  times  too  large  as  there  are  units  in  d. 
Therefore  the  product  of  a — b  by  d=ad — bd,  must  be  subtracted 
from  ac — be,  thus  {ac — be)  —  {ad — bd)=ac — be — ad-^bd ;  but 
-{-bd  is  the  product  of  — b  and  — d  ;  therefore  a  minus  quantity 
multiplied  by  a  minus  is  a  plus  quantity,  Q.E.D. 

80.  That  the  product  of  two  minus  quantities  produces  a 
plus,  may  be  illustrated  by  the  following  diagram  : 

Let   ABCD  be   a   right-angled   parallelogram.     Let  JH  be 

parallel  to  AB,  and  EG  parallel  to  AD.     Then  the  figure  will 

V  contain  four  right-angled  parallelograms,  JFGD,  AJFE,  EBHF, 

and  FHCG.     Let  AB,  which  is  equal  to  JH,=za,  and  EB  or  its 

3=^ 


30 


ALGEBRA 


F 
G 


equal  FH=b  ;  then  AE,  or  its  equal  JF,  will  be  :=^a—h.  Also 
let  J.D=c,  and  J.  7=^,  then  J  J)  or 
FG=c—d.  Now,  to  find  the  con- 
tents of  JFGD,  we  must  multiply  the 
adjacent  sides  of  the  parallelogram 
together,  which  are  JD  and  JF. 
ButJD=c—d,  and  JFz=za—b;  there- 
fore the  contents  of  the  parallelogram 
will  be  [a—h)y,{c—d)=zac—ad—bc-\-hd. 

But  ac  is  the  contents  of  the  figure  ABCD,  for  it  is  the  pro- 
duct of  the  adjacent  sides  AB  and  AD.  And  this  exceeds  the 
contents  of  the  figure  JFGD  by  the  three  parallelograms  AJFE, 
EBHF,  and  FHCG.  But  ad  is  the  contents  of  the  figure 
ABHJ,  for  the  side  AB=^a,  and  AJ=d,  and  these  are  the 
adjacent  sides  of  the  parallelogram.  And  be  is  the  contents  of 
the  figure  EBCG;  for  EB=b,  and  AD  or  BC=c,  and  there- 
fore bc=EBCG,  for  it  is  the  product  of  the  adjacent  sides  EB 
and  BC.  But  the  parallelograms  ABHJ  and  EBCG  both  in- 
clude the  parallelogram  EBHF;  whereas  it  should  be  included 
by  only  one  of  them.  It  must,  therefore,  be  returned.  The 
contents  of  this  figure  EBHF=bd ;  for  FH=b,  and  HB=d, 
and  their  product  is  bd.  And  as  BF  has  been  taken  twice  from 
the  figure,  it  is  restored  by  considering  bd  a  plus  quantity,  thus 
-\-bd,  Q.E.D. 


EXAMPLES. 

1.  Multiply  4:?n  by  'Sn. 

2.  Multiply  Sab  by  —bed. 
8.  Multiply  Sm7i  by  Axy. 

4.  Multiply  7pg  by  y. 

5.  Multiply  — ISadefhy  &m?ip. 

6.  Multiply  7hp  by  4tuz. 

7.  Multiply  19^3  by  — xyz. 

8.  Multiply  7 an  by  —2an. 

9.  Multiply  baaa  by  Saaa. 


A^is.  12mn. 

Ans.  — Ibabed. 

Ans.  S2vi7ixy. 

Ans.  7pgy. 

Ans.  — 7^adefmnp. 

Ans.  2Shpticz. 

Ans.  — Idabxyz. 

Ans.  — 14:aa7i?i. 

Ans.  Ibaaaaaa. 


MULTIPLICATION.  31 

10.  Multiply  — 4:xy  by  — xxyy. 

11.  Multiply  —  ll^fcby  —Sdee. 

12.  Multiply  97nn  by  2mmn. 

13.  Multiply  17 abc  by  —Sabc. 

14.  Multiply  Wxyyy  by  — yy. 

15.  Multiply  — 9mm?}i  by  — nn7i. 

16.  Multiply  ^2-?  by  pqt. 

8i.  When  the  same  letter  is  repeated  in  the  product,  for  the 
sake  of  brevity  one  letter  only  need  be  wi'itten,  with  a  figure 
placed  after  and  above  it,  denoting  the  number  of  times  the 
letter  is  taken  as  a  factor. 

This  figure  is  called  the  exponent  or  power  of  the  letter,  and 
it  shows  how  many  times  the  letter  is  used  as  a  factor.  Thus, 
(]^=.ay^ay^a=.(um^  4m'^=4XwX^=4ww. 

82i  If  two  or  more  letters  of  the  same  kind,  having  expo- 
nents, are  to  be  multiplied  together,  we  wi'ite  the  letter,  and 
place  over  it  the  sum  of  the  exponents.  Thus,  the  product  of  a? 
by  c^=zaaay^aa=aaaaa=a^.     Hence  the  following 

Rule.  Add  the  exponents  of  the  same  letter,  and  place  their 
sum  over  the  product  of  the  letter  multiplied  by  the  coefficients, 

17.  Multiply  Am^  by  3m-. 

4x  3Xw^'X?^'=12m^+'=12m^ 

18.  Multiply  —bn^  by  —An\  A?is.  20n\ 

19.  Multiply  —3a'"  by  oa'\  Ans.  —  9a^"'. 

20.  Multiply  2x"'  by  4:c".  Ans.  8^;'"+". 

21.  Multiply  Sa'b'  by  6a'b.  Ans.  lba'b\ 

22.  Multiply  ab""  by  a'b.  Ans.  a'P. 

23.  Multiply  a'b'^c  by  a'^bd.  Ans.  a'b'cd. 

24.  Multiply  7a^c^  by  a^cm.  Ans.  7aYm. 
^\b.  Multiply  M'b^x'  by  -a%^cx\  Ans.  -9a^%hx^\ 

26.  Multiply  15??iV  by  ^mn.  Ans.  4:5m^n\ 

27.  Multiply  Sa'^b"  by  2a'"3l  Ans.  Qa""'b''+\ 

28.  Multiply  4:X"'y''  by  — .^"yV-  ■^^^-   —  4a;'""^-"2/-V. 


32  ALGEBRA. 

29.  Multiply  lla^c^  by  4aacc.  Ans.  68aV. 

30.  Multiply  3^^"^+"  by  — 4a"^".  ^?w.   —12a'''". 

31.  Multiply  7a'"  by  3^-^  ^7i5.  21. 

32.  Multiply  ll?z^  by  —^n\  Ans.  — 55?il 

33.  Multiply  4a'  by  — 3^-2.  Ans.  —12a\ 

34.  Multiply  Tttz"  by  Sm\     .  Ans.  21m'". 

35.  Multiply  Qab''  by  a'^"*.  Ans.  Qa'b-\ 

36.  Multiply  a"'  by  a-^.  Ans.  a-\ 

37.  Multiply  rr""  by  a;\  ^    An^.  1. 

38.  Multiply  77i^  by  m~'^.  Ans.  1. 

Case  II. 

83.  When  the  multiplicand  is  a  compound  quantity,  and  the 
multiplier  is  a  simple  quantity. 

Rule.  Multiply  each  term  of  the  multiplicand  separately  by 
the  multiplier.,  and  prefix  the  proper  sign  to  each  term  of  the 
product. 

EXAMPLES. 

(1)                   (2)                   (3)  (4) 

Multiply  ^a-{-bx           Im—^n           3^— 4c  bx-\-lb 

By      Am                 Sa                   be  3m 


12am-{-20?nx.  2\am — 12a7i.  Ibbe — 20ce.  lb7nx-\-21bm. 

(5)  (^^)  (7)  (8) 

Ax'—Zax'      Am^-\-2n  Mbc—d  abc-\-7n" 

Aam 


12x^—9ax^  '[2m'-\-6m'n.   AOa'bcd'—bad\  Ad'bcm-{-Aa7n"+\ 

9.  Multiply  ba'x—ly+Ax^—W'  by  4:ay\ 

Ans.  20a%--28a?/^+16aa;y — 1 2ajy. 

10.  Multiply  Id'b^-^Aam'—Qy  by  4aW. 

Ans.  2'^a'bV-{-lQ,a'm-'—24:a'mhj. 

11.  Multiply  4a'^i^— 6a''c+c*^  by  —ba\ 

Ans.   — 20a'^»^4-30a'c-5aV. 


MULTIPLICATION.  33 

12.  Multiply  —ab'^—dx^—Hm-'^hj  —ain. 

Case  III. 

84.  When  both  the  multiplicand  and  multiplier  are  com- 
pound quantities. 

Rule.  Multiply  each  term  of  the  multiplicand  by  each  term 
of  the  multiplie)\  remembering  that  tJw  product  of  like  signs  is 
-f-,  a7id  the  product  of  unlike  signs  is  —  ;  then  add  together  all 
the  products. 

Note.    Terms  wMch  are  alike  should  be  placed  under  one  another. 
EXAMPLES.  ^' 

(1)  (2) 

Multiply      3a-{-4:b  x-{-y 

By  2a+  b  2x—y 


Qd'+Sab  2x'-{-2xy 

-\-Sab-\-4:b'^  —xy—y 


%d'-[-llab-{-4:b\  2x^-]-xy—7/ 


(3) 

lax—4iy-\-Qm 


/    ^      •^- 


4«2/+%  k    Y-i     -    -^^ 


f*^ 


2Mxy—\Qay'+2^amy  if-     >;    "^  ^  ^^ 

-\-14:axy — Sy'^~\-1277iy 

2Sa^xy—lQay'^-\-14:axy—'8y'^-\-24:amy-{-12my. 

(4)  (5)  (6) 

2x''-{-y  Sa-\-A7n  Sa—2b 

x~-\-y  2a — 27n  2a — bb 


2x'-]-7?y  6«2+8a7?i  6^2—4^3 

\.2x'y-{-y''  _6a?7i— 8?;i^  —15^3+1031 

2x'-\-^x'y-\-7f.  6a^-{-2am-Sm\  Qa''—ldab-\-10b\ 


34  ALGEBRA. 

(7) 
20"— Za  b—4  b'' 


—12a'b-\-lba%''-\-24:d'b'-^ab' 

—lQa'b"-\-2^a%^-\-Z2ab'—U^ 

U'—22a'b—^la?h'-\-'i'^(rb'-{-2Qab'—U\ 

85.  When  positive  and  negative  terms  balance  each  other  in 
the  product,  they  should  be  cancelled. 

(8)  (9) 

a—  X  1  -{-X 

a^-{-a^x-{-ax'^  l—x-]-x'^—x^ 

—a^x—ax^—x^  -\-x—x^-\-x^—x'^ 

a?  —x^.  1  —x\ 

86,  The  continued  product  of  factors  is  often  expressed  in 
one  line. 

10.  (1+2-)  {l+x')  (l—x-{-x''—x'')=l—x\ 

11.  {a-\-2x)  {a—3x)  {a-\-4:x)=a'-\-Sa'x—10ax''—24:x\ 

12.  Required  the  continued  product  of  oa—x,  2a-^4x,  and 
4:a-2x.  Am.  24:a'-\-2Sa:x—^Qax''+^xK 

13.  Multiply  2>x^—2xy—y'  by  2x—^. 

Ans.  6z^ — lQx'^y-\-'Qxif-\-^y^. 

14.  Multiply  x^-\-2x-\-l  by  x'—2x-\-^. 

Ans.  2:^-1-4.^+3. 

15.  Multiply  a-\-b — c  by  a — b-\-c. 

Ans.  a'—b''-\-2bc—c\ 

16.  Multiply  ^a—2b  by  —2a-\-^b. 

Ans.  — 6a*^+16a3— 851 
\    17.  Multiply  ba'-'^ab-\-W  by  6^-53. 

A71S.  30a^—43a'5+39«^.=— 20^*1 
18.  Multiply  a'-\-ab-\-b''  by  a-b.  Ans.  a?—b\ 


MULTIPLICATION.  35 

19.  Multiply  o^—x"  by  o>—x\  Ans.  a^—2a'x'-\-x\ 

20.  Multiply  'Ix^'—Sxy-^-Q  by  3x^+3a:?/— 5. 

Ans.  6x'—dx'y-\-Sx-—9xy-i'^^xy—^0. 

■     21.  Multiply  ba'—4:ax-\-Sx''  by  2a'—Sax—4:X-. 

Ans.  10a'-2Sa'x-2orx''+7ax'—12x\ 

^  22.  Multiply  2a'—Sax^4:x"  by  ba'—Qax—2x\ 

Am.  10a'-27a'x-\-S4:a'x'—l^ax'—Sx\ 

23.  Multiply  a'-Sa'+Sa-l  by  a'^_2«-|-l. 

^"■""     ^7W.  a'-ba'+lOa'-lOa'+ba-l. 

24.  Multiply  «'"—«''  by  2a—a'\ 

Am.  2a"^^—2a"+'—a'"+''-\-a^\ 

25.  Multiply  a^—a^x-\-a^x^—ax^~^x^  by  o;-}-^- 

MULTIPLICATION    BY   DETACHED    COEFFICIENTS. 

8T.  The  coefficients  of  the  polynomials  should  be  arranged 
according  to  the  successive  powers  of  the  letters,  increasing  or 
decreasing  by  a  common  diflFerence  ;  and,  when  this  common 
difference  is  wanting,  its  place  should  be  supplied  by  zero. 

The  following  examples  will  illustrate  the  above  : 

1.  Multiply  a'-{-2a+l  by  a''—2a-{-l. 
1+2+1 
1-2+1 


1+2+1 
_2-4-2 

+1+2+1 

1+0-2+0+1. 

In  adding  the  coefficients  of  the  partial  products,  we  perceive 
that  the  second  and  fourth  places  are  a  zero  :  but  the  letters 
must  be  written  with  their  powers  regularly  ascending  from  left 
to  right ;  and,  where  zero  is  the  coefficient,  the  value  of  the 
quantity  is  nothing.  Thus,  a'-i-0a^—2ar~{-0a-{-l=a'—2a'-{-l, 
because  zero  is  the  coefficient  of  the  second  and  fourth  terras. 


36  ALGEBRA, 

2.  Multiply  x'—x"  by  x^-\-x. 
1+0-1 
1+0+1 

1+0-1 

_|_l+0-l 

1+0+0+0-1. 
With  the  letters  and  their  powers  added,  it  will  be 

The  second,  third,  and  fourth  terms  are  of  no  value. 

8.  Multiply  3^3- 4a3H6^'by  1d?—U\ 
3+0-4+6 

2+0-4 

6+0—  8  +  12 

-12-  0+16-24 


6+0-20+12+16-24. 

We  now  annex  the  letters  with  their  proper  powers,  decreas- 
ing by  a  constant  common  difference,  thus : 

6a^— 20«^^^+12a^3^+16a3^— 243'. 

4.  Multiply  2a?-^ab''-\-bb^  by  2a^-U\ 
2+0—  3+  5 
2+0-  5 


4+0—  6+10 

_10_  0  +  15-25 

44_0— 16+10+15— 25. 

Affixing  the  letters  with  their  powers,  we  have, 

^a^-\-^a'b—l^a'b''-\-10a'b^-{-lhab'—'lbb'= 

5.  Multiply  5a'— 3a^+^  by  2a'-{-a\ 

Ans.  10fi^9+5a'— 6a'— a'+a\ 

6.  Multiply  3.t3-2.7:-2  by  2:^-3. 

Ans.  3:r'-lla:^— 22'-+6a:+6. 


DIVISION.  37 

7.  Multiply  ?/^-f-y— 3  by  if—y. 

Arts.  y^J^y^-^f-o/J^^. 

8.  Multiply  x''-{-x''-{-x^-\-z'-\-x-\-l  by  x—l. 

Ans.     x^ — 1. 

9.  Multiply  a"^— 2a34-452  by  d}-^1db-\-W, 

Ans.  a'^^a%''-\r\U\ 

10.  Multiply  3a'+3a'3+3^23'^+3a^^+33^  by  7a— 7^. 

A.m.  21a'— 2W. 

11.  Multiply  x^-}-x-y-\-xif-^7^  by  x — y.  Ans.  x^ — y^. 


SECTION    V. 

DIVISION. 

Art.  88:  Division  is  the  converse  of  Multiplication,  and  is 
performed  like  that  of  numbers.  Its  object  is  to  find  how 
many  times  one  quantity  is  contained  in  another;  or  to  find 
what  quantity,  multiplied  by  a  given  quantity,  will  produce 
another  given  quantity. 

The  product  of  like  signs,  as  in  the  rule  of  Multiplication, 
produces  +,  and  unlike  signs  — . 

Case  I. 

89t  When  the  divisor  and  dividend  are  both  simple  quan- 
tities. 

If  dhc  be  divided  by  a,  the  quotient  will  be  he  ;  because  a 
multiplied  by  he  will  produce  ahc. 

If  ^ahc  be  divided  by  2a^  the  quotient  will  be  2hc  ;  because  2a 
multiplied  by  2hc  will  produce  4(z3c. 

If  9^:r  be  divided  by  3a;,  the  quotient  is  33  ;  for  33  multiplied 
by  %x  is  ^hx. 

From  the  above  illustration  we  derive  the  following 

Rule.     Write  the  divid.end  over  the  divisor,  in  the  manner  of 
4 


38  ALGEBRA. 

a  fraction^  aiid  reduce  it  to  its  simplest  form  by  cancelling  the 
letters  and  figures  tJiat  are  common  to  all  the  terms. 

Or,  divide  the  coefficient  of  the  dividend  hy  the  coefficient  of 
the  divisor,  and  cancel  the  letters  common  to  the  divisor  and 
dividend. 


EXAMPLES. 


1.  Divide  ^ab  by  2a. 


^~=Sb  ;  or,  Qab-^2a=Sb. 
za 

2.  Divide  12abcxy  by  4:bx. 

.^  — — — -=Sacy  ;  or,  'i.2abcxy-r-4:bx=3acy. 

3.  Divide  m?iop  by  op.  Arts.  mn. 

4.  Divide  7(z5??z  by  a7?i.  Ans.  lb. 

5.  Divide  14:xyz  by  7x.  An^.  2yz. 

6.  Divide  lOabcd  by  ^bcd. 

7.  Divide  9mnx  by  ox. 

8.  Divide  17 ab  by  ab. 

9.  Divide  Adqrst  by  7qt. 

10.  Divide  20hm7io  by  4:?io. 

90.  Powers  and  roots  of  the  same  quantity  are  divided  by 
subtracting  tbe  index  of  the  divisor  from  that  of  the  dividend. 
Thus,  if  we  wish  to  divide  a^  by  a^,  we  subtract  the  index  3 
from  the  index  5,  and  set  the  remainder  2  over  the  a;  thus,  a^. 
This  process  is  evident  from  the  fact  that  a^=aaaaa,  and  a^ 
=aaa,  and  aaaaa  divided  by  aaa  gives  0^=0^. 

11.  Divide  4:a'b*  by  2ab\ 

-^=2a'^^%-  or,  AaW-i-2ab''=2a'b\ 
Aa¥ 

12.  Divide  7a^  by  (r.  Am.  7a\ 

13.  Divide  Ga'b'cd  by  Sab.  Ans.  2a%cd. 

14.  Divide  Try  by  r^.  An^.  7r'^p^. 

15.  Divide  60/?/'  by  30/.  Ans.  2pY. 

16.  Divide  \2axY  by  4^2:1  Ans.  3?/. 

17.  Divide  96r\vfV  by  48.<ffV.  Ans.  2r*2t\ 


DIVISION.  39 

18.  Divide  VJa^xif  by  17.  Arts.  a^xy'. 

19.  Divide  aP  by  a^.  Ans.  a^. 

Case  II. 

91 1  When  the  divisor  is  a  simjjle  quantity,  and  the  dividend 
a  compound  one,  we  adopt  the  following 

Rule.     Divide  each  term  of  the  dividend  by  tlie  divisor ^  as  in 
Art.  89.     Or,  v)e  may  write  the  divisor  under  the  dividend,  in 
the  form  of  a  fraction,  and  then  cancel  equal  quantities  when 
found  in  the  divisoi'  and  in  each  term  of  the  dividend. . 

EXAMPLES. 

1.  Divide  ^a^b+Qa'c—l2ah  by  3a. 

OPERATION. 

Za)^a'b-\-Qa'c—12ah 


U%-\-2ah—  U.  Am. 
We  find  that  3a  is  a  factor  in   each  term  of  the  dividend ; 
we   therefore   write   the   other  factors   under   their   respective 
quantities. 

2.  Divide  8a^^c-f  16a^^c-4aV  by  ^d'c. 

Ans.  2ab-\-4:a^b—c. 

3.  Divide  da'bc—3a''b-\-lSa^bc  by  Sab. 

Ans.  ?>a^c — a-\-Qah. 

4.  Divide  1()a'bc-\-lbabd'—l^a^be  by  bab. 

Ans.  4a-c-{-3d^—2ae. 

5.  Divide  15x^+30^^2/^  ^J  ^^'  -^^• 

6.  Divide  lax'^yz^ — 14iXyz-\-21xy'^  by  Ixy.  Ans. 

7.  Divide  p^mq-\-p^m — p'^mc  by  p^.  Ans. 

8.  Divide  Atxz—St'^z+z'^  by  z.  Ans. 

9.  Divide  12a-''~Sa'b-\-lQa'x-10a-^y  by  2a\ 

Atis.  Ga-^ — 4:b-\-Sax—ba~*y. 

Case  III. 

92.  When  the  divisor  and  dividend  are  both  compound  quan- 
tities. 


40  ALGEBRA. 

Rule.  Write  down  the  quantities  in  the  same  manner  as  in 
the  division  of  numbers  in  Arithmetic,  arranging  the  terms  of 
each  quantity  so  that  the  highest  powers  of  one  of  the  letters  may 
ftand  before  the  next  loicer. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor,  and  set  the  result  in  the  quotient,  ivith  its  proper  sign. 

Multiply  the  whole  divisor  by  the  terrfythus  found  ;  and,  having 
subtracted  the  result  from  the  dividend,  bring  down  as  manij 
terms  to  the  remainder  as  are  requisite  for  the  next  operation, 
which  perfm'm  as  before  ;  and  so  proceed.,  as  in  Arithmetic,  till 
the  work  is  finished. 

1.  Divide  a^-^'lah^W  by  a^b. 

2  I  o  I  r  7.2/^+^  divisor. 
a^-\-1ab-\-b\  — pj- 

\a-\-b  quotient. 

c^-\-  ab 

ab^b"" 
ab-\-W 

2.  Divide  c^-\-^a}x-{-hax^-\-x?  by  a-{-x. 

a-\-x 


a^4-ba^x4-bax'^-\-x^l  —r 
\a' 


'-{-4:ax-{-x\ 
a^-\-a^x 


^a^x-\-bax^ 
Aa^x-^-^ax^ 


ax^-{-x^ 
ax'^-\-x^. 

3.  Divide  a'-\-^a'b''-\-lW  by  a'—2ab-^^b\ 

a'-2a%-]-4.a~P 

2a^b-\-lW 
^a^b—^a^b^-^'^ab^ 


4:a%''—Ub^-\-l^b' 


DIVISION.  41 

It  may  be  verified  that  a^-^2ab-{-4:b'^  is  the  true  quotient,  by 
multiplying  it  by  the  divisor.  It  should  also  be  observed,  that 
in  every  stage  of  the  proceeding,  the  terms  involving  the  highest 
powers  of  a  have  been  placed  first  on  the  left. 

4.  Divide  4x'—9a'x'-{-Qa'x-a'  by  2x''—^ax-{-a\ 

Ax'—Qax^-\-2a''x'' 

Qax^ — lla^x^'-^-Qa^x 
Qax^ — da^x~  -\-  Zcc'x 


9Si  If  the  divisor  be  not  exactly  contained  in  the  dividend, 
the  quantity  that  remains  after  the  division  is  finished  must  be 
placed  over  the  divisor  at  the  right  of  the  quotient,  in  the  form 
of  a  fraction. 

5.  Divide  c^ — x^  by  a-\-x. 

a-{-x 


'-x'f- 


^a^—ax-\-x'^- 
a^-{-a^x 


2x 


a-\-x 


—d^x- 
— drx- 

-x' 
-ax^ 

ax^—x^ 
ax'^-\-x^ 

-2x\ 

Oil  The  operation  of  division  may  be  considered  as  ter- 
minated when  the  highest  power  of  the  letter,  in  the  first  or 
leading  term  of  the  remainder,  is  less  than  the  first  term  of  the 
divisor. 

The  division  of  quantities  may  also  be  sometimes  carried  on 
ad  infinitum,  like  a  decimal  fraction ;  in  which  case  a  few  of 
the  leading  terms  of  the  quotient  will,  generally,  be  sufficient  to 
4# 


42  ALGEBRA. 

indicate  the  rest,  without  its  being  necessary  to  continue  the 
operation. 

6.  Divide  a  by  a-\'X. 

(        X    x'    x' 


a     a"     a 


a~\-x 


— X 

x" 

— X 

a 


a 

x^ 

a' 

x^ 

x' 

a?' 

~a?' 

7.  Divide  a  by  a — x. 

8.  Let  c^ — 2ax-{-x^  be  divided  by  a — x.  Ayis.  a — x. 

9.  Divide  a?—2,d'h^Zab''—b^  by  a-b. 

Ans.  a^—lab^y. 

10.  Divide  'ia^-^a-b-^alJ'-^-W  by  "la-h. 

Am.  ^a^—Zh^. 

11.  Divide  Zb^-\-ZaW—^a%—^a^  by  a^b. 

Ans.  —^a^J^W. 

12.  Let  2aV— Sdiz-f  2  be  divided  by  "lax—l. 

Ans.  ax — 2. 

13.  Divide  Ila'-'IW  by  la-lb. 

Ans.  Za'-\-Za^b-\-Za%^-\-Ub^+U\ 

14.  Divide  x'-y'-\-2y''z'—z'  by  x'-\'y''—z\ 

Ans.  x"^ — y'^-^-^^- 


DIVISION.  43 

15.  Divide  l-\-a  by  1—a. 

Am.  l-|-2a4-2a2+2a^-f  2a^+,  &c. 

16.  Divide  Sx''—lby''-\-2dyz—2xy—Sxz—Qz''hj2x—Sy+z, 

Ans.  4:X-\-by — Q>z. 

17.  Divide  6a;^— 96  by  3a:— 6. 

A7U.  2z'+4z'+8^-fl6. 

18.  Divide  a^J^a%''-{-a'b'-\-a'b'+b^  by  a'^a^b+a?b''-\-ab^ 
+b\  Ans.  a'—a%^a'b''—ab^-\-b\ 

DIVISION   BY   DETACHED   COEFFICIENTS. 

95 1  As  the  pupil  has  seen  in  Art.  87  that  the  operation  of 
many  questions  in  Multiplication  is  facilitated  by  using  de- 
tached coefficients,  he  will  readily  perceive  that  the  same  prin- 
ciple will  apply  to  Division. 

The  terms  of  the  divisor  and  dividend  are  to  be  arranged 
according  to  the  power  of  the  letters,  and  zero  must  be  inserted 
in  the  terms  that  are  wanting. 

The  first  literal  term  of  the  quotient  is  obtained  by  dividing 
the  first  letter  of  the  dividend  by  the  first  letter  of  the  divisor ; 
and  the  letters  belonging  to  the  other  terms  are  written  in  the 
same  order,  as  they  are  found  in  the  divisor  and  dividend. 

EXAMPLES. 

1.  Divide  a3+3a'3+3a32+^»3  by  a^b. 

/    l+l      coefficients  of  the  divisor. 
\l-|-2-f-l  coefficients  of  the  quotient. 
1+1 

2+3 

2+2 


1+1 
1+1. 

a^-7-f^=(2^,  first  literal  term  of  the  quotient.  The  others  will 
therefore  be  a3+3^  and  these  terms  annexed  to  the  coefficients 
will  be  a-+2«5+3l 


44  ALGEBRA. 

2.  Divide  x^ — if  by  x~ — y^. 

1+0-1 


l+0+0+0-l(-^ 
1+0-1 


1+0-1 
1+0-1. 

x''-^x^=.x^^  first  literal  part.  The  other  regular  parts  are 
xy-\-y^.  Having  prefixed  the  coefficients,  it  will  be  x'^-\-Oxy 
+2/^;  but,  as  the  coefficient  of  the  second  term  is  zero,  the 
term  has  no  value.     The  correct  answer  will  therefore  be  x^-\-y'^. 

3.  Divide  3a;^— 48  by  3a:— 6. 

3+0+0+0-48(j^- 
3-6 


6 
6-12 


12 

12-24 


24-48 
24-48. 

x'^-^x=.x'^,  first  literal  part.  The  succeeding  terms  will,  there- 
fore, be  x^-{-x-\-x^.  Hence  the  true  quotient  will  be,  x^-\-2x^-\- 
4a;+8. 

4.  Divide  \—a?  by  l-\-a. 

Ans.  1 — a-{-(r — a?-\-a'^ — a^-\-a^ — (^. 

5.  Divide  3?/^+3a:?/^ — ^x^y—^x^  by  x-\-y.     A7is.  oy'^ — 4:X~. 

6.  Divide  a'-^a'b-Sa'b'+l^aP-Sb'  by  a"-i-2ab-2b\ 

Am.  a'^—5ab-\-4:b\ 

7.  Divide  m^—b77i'n-^107n^n~—10m'^n^-{-5?n?i^—n^  by  tti^ — 
2m7i-{-7i-.  Atis.  m^ — 2,7ri?n-\-omnr — n^. 

8.  Divide  a''H-"_|_a"+i^'«-i__a"'->Z,''+i_^"»+"  by  ^"-^+3"^^ 

Ans.  a"+i— 3"+^ 


MISCELLANEOUS     QUESTIONS.  45 

QUESTIONS    TO   EXERCISE    THE   FOREGOING   RULES. 

1.  What  is  the  sura  of  the  following  quantities :  12a-|-5c-j- 
17^-1-133,  Sa-\-12b-^lbd-^8c,  llc  +  lba-l-2Sb-^10d,  and  U-f- 
3«-|-20Z»+18c  ?  A7i^.  38a+683+42c+46rf. 

2.  Add  together  5(Z-|-35— 4c,  2a—bb-^(5c-}-2d,  a— 4b— 2c 
-]-3e,  and  7«+43— 3c— 6e.  A?is.  15a— 2^— 3c-[-2^— 3c. 

8.  Find  the  sum  of  3a2+2ai+4^-,  bar—^ab-\-Qb\  —4a:-{-bab 
—b\  lSa:'—20ab—l%\  Ua'—2ab-\-20b\  and  —^Qa:'-\-24:ab— 
10b'.  Ans. 

4.  Kequired  the  sum  of  5a^5— ITa^^c- 15^V-[-5,  —4:ab-{- 
8a;'bc—10bY—4:,  —^a'b—Sa'bc-\-20b'c'—S,  and  2arb-\-12a'bc-\- 
53V4-2.  ^7^5. 

5.  Add  the  following  quantities:  a-^b-\-c-{-d,  a-\-b-\-c — 2c?, 
a-\-b—2c-{-d,  a—ob-{-c-{-d,  —a-\-b-\-c-{-d,  and  a— b— 2c— 2d. 

Ans.  4:a. 

6.  Multiply  x'-\-2x-{-l  by  x'—2x-\-S.       Ans.  x'-]-4:X^S. 

7.  Multiply  l—x-\-x''—x^  by  1+a:.  A7is.  l—x\ 

8.  Multiply  l_2a;+3a;^— 4^'3+5:c^-6:i'^+72;^-82-^  by  1-f 
2x-\-x''.  Ans.  l—9x^—Sx'. 

9.  What  is  the  continued  product  of  (i-{-b,  a — b,  a'^-\-ab-{-b'^, 
and  a^—ab+P  ?  Ans.  a'—b\ 

10.  Multiply  x'-Jroax"+^a'x-i-a^  by  x'—dax''+8arx—a\ 

Ans.  x'—dd'x'-\-Sa'x''—a\ 

11.  Multiply  a^'-'-^b'"-'  by  «"+'— 3"+^ 

J[?Z5.  a'"+"-\-a''+^b"'-^—a'^-^b"+'^—b'"-^". 

12.  Divide  rc^ — a^  by  a; — a.    Atis.  x'^-\-ax^-\-d^x^-\-a?x-\-a^. 

13.  Divide  a:* — ^x^ — ^xy — if'  by  x^-\-'^x-\-y. 

Alls,  rr^ — ox — y. 

14.  Divide  x^—4:X^-\-Qx'^—4x-\-'l  by  a;^— 2:r-{-l,  and  x^— 
2a^x^-\-lQia^x — Iba^  by  x^-\-2ax — 3^^^,  and  find  the  difference  of 
their  quotients.  Ans.  2x — 2ax — l-}-5a^ 

15.  Divide  a;^— 16aV4-64a^  by  x—2a. 

Ans.  2;^+2az*+4aV— Sa^:^-— IGa^a;— 32al 


46  ALGEBRA. 

SECTION    VI. 

FRACTIONS. 

Aet.  96.  Algebraic  Fractions  are  similar  to  vulgar  frac- 
tions in  Arithmetic ;  they  express  a  part,  or  parts,  of  a  quantity 
or  a  unit. 

D7o  They  consist  of  two  parts,  the  numerator  and  denom- 
inator, the  former  being  written  above  the  line,  and  the  latter 
beloiv  it ;  and  these,  when  taken  together,  are  the  ter^ns  of  the 
fraction. 

98 1  The  denominator  shows  into  how  many  parts  the  quan- 
tity or  unit  is  divided ;  and  the  numerator,  how  many  of  these 
parts  are  represented  by  the  fraction. 

99.  A  proper  fraction  is  one  whose  numerator  is  less  than 
its  denominator ;  as, 

a—h        7 
a-\-d'  """^  8* 

100.  An  hnpi'oper  fraction  is  one  whose  numerator  is  equal 
to  or  greater  than  its  denominator ;  as, 

a  b-\-c  7 
-,  or  — - ,  or  K. 
a        0 — c        6 

101.  A  mixed  quantity  is  a  whole  number  or  quantity,  with  a 
fraction  annexed,  with  the  sign  either  plus  or  minus ;  as, 

a  m  ,  a  m       ^„ 

~-\-y,  or X,  or  y-\ — ,  or  x ,  or  7f. 

b  '  ^        n  -^  '  c  n  ^ 

102.  A  compound  fraction  is  a  fraction  of  a  fraction;  as, 

-  of-  of-  ;  or,  I  off  of -^V 

103.  A  complex  fraction  is  a  fraction  having  a  fraction  in  its 
numerator  or  denominator,  or  in  both :  as. 


FRACTIONS. 

3 

4 

a            a 

4 

7 

b          c — d 

14X' 

11' 

-,   or    

c              7^ 

12 

^            p 

47 


104.  The  value  of  a  fraction  depends  on  the  ratio  which  the 
numerator  bears  to  the  denominator. 

105.  The  value  of  a  fraction  is  not  changed  by  multiplying  or 
dividing  both  numerator  and  denominator  by  the  same  quantity. 

106.  The  greatest  common  measure  of  two  or  more  quantities 
is  the  largest  quantity  that  will  divide  all  of  them  without  a 
remainder. 

107.  The  least  common  multiple  of  two  or  more  quantities  is 
the  least  quantity  that  can  be  divided  by  them  all  without  a 
remainder. 

108.  A  fraction  is  in  its  lowest  terms  when  no  quantity,  ex- 
cepting a  unit,  will  divide  both  of  its  terms. 

109.  Quantities  are  said  to  be  prim£.  to  one  another  when 
their  greatest  common  measure  is  a  unit. 

110.  Frime  factors  0^  quantities  are  those  factors  which  can 
be  divided  by  no  quantity  but  themselves  or  a  unit ;  thus,  the 
prime  factors  of  35  are  7  and  5. 

111.  A  composite  quantity  is  that  produced  by  multiplying 
two  or  more  quantities  together. 

112.  A  fraction  is,  in  value,  equal  to  the  number  of  times  the 
numerator  contains  the  denominator. 

113.  A  fraction  is  increased  in  value  either  by  multiplying  its 
numerator  or  dividing  its  denominator. 

114.  A  fraction  is  diminished  in  value  either  by  dividing  its 
numerator  or  multiplying  its  denominator. 


48  ALGEBRA. 


Case  I. 

115.  To  find  the  greatest  common  measure  or  divisor  of  the 
terms  of  a  fraction. 

Rule.  Arrange  the  two  quantities  according  to  the  order  of 
their  powers,  and  divide  that  which  is  of  the  highest  dimensions 
by  the  other,  having  first  cancelled  any  factor  that  may  be  con- 
tained in  all  the  terms  of  the  divisor,  without  being  common  to 
thnse  of  the  dividend. 

Divide  this  divisor  by  the  remainder,  simplified  as  before,  and 
so  on  for  each  successive  remainder,  and  its  preceding  divisor, 
till  nothing  remains ;  and  the  last  divisor  will  be  the  greatest 
common  measure  or  divisor  required. 

If  any  of  the  divisors,  in  the  course  of  the  operation,  becoTne 
negative,  they  m.ay  have  their  signs  changed,  or  be  taken  affirm- 
atively, without  altering  the  truth  of  the  result ;  and,  if  the  first 
term  of  a  divisor  should  not  be  exactly  contained  in  the  first  term 
of  the  dividend,  the  several  terms  of  the  latter  may  be  midtiplied 
by  any  number  or  quantity  that  loill  render  the  division  com- 
plete. ^ 


EXAMPLES. 


CX-\-X^ 


1.  Find  the  greatest  common  measure  or  divisor  of -5 — ; — ^. 
°  ac-\-a^x 

cx-\-x^)a^c-{-(i^x 

c-\-x)  a^c-{-a^x{cr 

ac-\-a~x. 

As  X  is  found  in  both  terms  of  the  divisor,  we  divide  those 
terms  by  x  before  the  operation. 

The  greatest  common  measure  of  both  terms  we  perceive  is 
c-\-x ;   that  is,  it  will  divide  them  both  without  a  remainder. 

Thus,  c-{-x)  .^        ^  z=- 
ac-j-ax     a 


2.  Required  the  greatest  factor  of 


FRACTIONS.  49 


x'yilx'^lrx 

—2bx'—2b''x) 
x-\-b)x^-]-2bx-\-b\x-\-b. 
x'^-^-  bx 

bx-\-h' 
bx-i-b'' 
We  cancel  2bx  in  both  terms  of  the  second  divisor,  as  it  is 
common  to  both. 

As  x-\-b  is  the  last  divisor,  it  is  the  greatest  factor  or  com- 
mon measure  of  the  quantities  proposed. 

3.  Required  the  greatest  common  divisor  of  Sa^ — 2a — 1,  and 
4a3_2a-— 3«-}-l. 
3a2_2a— l)4a'— 2fl'^— 3a+l(4a 

o 
O 


12a'-Sar—4:a 

2a'—ba-{-^y6a~—2a-l 

2 

Qa"—  4a— 2(3 
6a2_15a4-9 

lla-11 

a-l)2d'—5a-{-^2a- 
2a'-2a 

-3 

—  3a4-3 
— 3a+3. 

As  11  is  common  to  both  terms  of  the  third  divisor,  it  is 
cancelled;  therefore  a—1  is  the  greatest  common  factor  of  both 
quantities. 

4.  "What  is  the  greatest  common  divisor  of  x^—a^,  and 
x"^ — a~  ?  ,'  _  o:  A?is.  X — a. 


5 


50  ^    —  Y^i  ALGEBRA. 

5.  What  is  tlie  greatest  common  factor  o^ x^ — 1,  and  ax-\-a? 

Ans.  x-\-\. 

6.  Required  the  greatest  common  factor  of?/"' — a;^  and  'if — 
ifx — yx^-\-x^.  Ans.  if — x\ 

7.  Required  the  greatest  common  measure  of  a^ — crx-{-ax'^ 
—x^^  and  a^ — x^.  Ans.  d^ — (j^x-\-a'Ji? — x^. 

8.  Required  the  greatest  common  factor  of  a'' — x^^  and  (jC'-\- 
a?x^.  ,/  ^^  '  '       Arts.  d^-{-x^. 

Case  II. 

116.  To  reduce  fractions  to  their  lowest  terms. 

Rule.  Divide  the  terms  of  the  fraction  by  the  prime  factors 
common  to  both. 

Or,  divide  both  terms  of  the  fraction  by  their  greatest  commoii 
divisor. 

IIT.  That  fractions  after  reduction  have  the  same  value  as 
before,  is  evident  from  the  fact  that  their  numerators  retain 
the  same  ratio  to  their  denominators;  for  equi-multiples  and 
sub-multiples  of  any  two  numbers  have  the  same  ratio  to  each 
other  as  the  numbers  themselves. 

Letters  or  numbers  common  to  all  the  quantities  in  each  term 
of  the  fraction  may  be  cancelled. 

EXAMPLES. 

^    ^   ,         ^abc       .     . 

1.  Reduce  tttt-.  to  its  lowest  terms. 

ba^bd 

4:abc       2abX^c       2c 

— -2—  z= .     Ans. 

Qd^bd     2abX.Sad     oad' 

In  this  operation  we  find  2ab  to  be  the  largest  factor  in  both 

2c 
terms ;  it,  therefore,  may  be  cancelled,  and  the  answer  is  ^—z. 

2.  Reduce  —, — —  to  its  lowest  terms. 

admny 

abxv         bx       . 

^=r  - — .  Alls. 

admny     dmn 


FRACTIONS.  51 

In  this  question  we  find  a  and  y  common  to  both  terms ;  and, 

hx 

they  beina;  cancelled,  the  result  is  -; . 

dmm 

3.  Reduce ^—  to  its  lowest  terms.  Ans.    — . 

mTwp-qx  px 

4.  Eeduce  ,  .,,.,  to  its  lowest  terms.  Ans.    ■=-. 

1 2(2771  4^3 

5.  Reduce  r^r- —  to  its  lowest  terms.  A71S.   -^r^. 

l^bcm  DOC 

6.  Reduce  ^^  „  „  to  its  lowest  terms.  Atis.    -—  . 

oQd'x^  9ax 

_,    _,'           QaTi'        .     .  .an 

7.  Reduce  —-7--  to  its  lowest  terms.  A71S.    -prr- 

obo7i  bo 

„    ^    ^        bQbTTi^x^-       .     -            ,  .         14:mx^ 

8.  Reduce  -77^ — 5-  to  its  lowest  terms.  Atis.    —-7- — . 

ibb7nxr  11) 

^    _,    ,         V^ah-cde^       .     .  a            b 

9.  Reduce  _,.  .,.    ,  -  to  its  lowest  terms.  Ans. 


Iba^cde  4aV 

10.  Reduce    „  ,  ^■, — — -,  to  its  lowest  terms. 
X  -\-YJ)X-\-b- 

In  performing  this  question,  we  first  find  the  greatest  common 
measure  of  the  two  terms  of  the  fraction,  which  is  x-\-b ;  wo 
then  divide  both  terms  by  it.     Thus, 

,  A     x^ — Wx         x~ — hx       . 
^     J  x'-^2bx~\-b^       x-\-b 

-.-.     -r.    1        Qa^-{-bax — Qx^         .     , 

■  11.  Reduce  ^.  ,  '     ^ — ^,  to  its  lowest  terms. 

ba^-\-loax-{-ox' 

Sa,—2x 
^a-\-'2x' 

12.  Reduce  — — —,  to  its  lowest  terms.  .  1 

a — x^  Ans 


x^ if 

13.  It  is  required  to  reduce  —. — ^  to  its  lowest  terms. 

x^ — y^ 


Ans. 


d'-\-x^' 
x^-\-xhf-\-y'^ 


x'-\-y\ 


52  ALGEBRA. 

Case  III. 

118i  To  reduce  a  mixed  quantity  to  the  form  of  a  fraction. 

Rule.  Multiply  the  integral  part  by  the  denominator  of  the 
fractional  part ;  to  this  product  annex  the  numerator  of  the 
fraction,  prefixing  to  it  the  sign  of  the  fraction  ;  under  the  whole 
write  the  denominatoi'  of  the  fraction. 

EXAMPLES. 

1.  Reduce  71-  to  a  fractional  form.        -^  =-^.     Ans. 

^  5  5 

2.  Reduce  a-\ —  to  the  form  of  a  fraction. 

e 

ayCe-\-b     ae-4-b 

'    = — —.     Ans. 


e  e 


3.  Change  a-{ to  a  fraction. 


ayCm4-a~ — b'     am-\-a- — b^ 

7n  rti 

7Yi-\—n 

4.  Chano;e  a ^^—  to  the  form  of  a  fraction. 

e 


ayce — m-\-n     ae — m — n 

-— ^ —  = .     Ans. 


7 

5.  Reduce  x to  the  form  of  a  fraction. 

m 


x'yCra — a — b     mx — a-\-b        . 
= .     Ans. 


171  m 

b^—cd 

6.  Chancre  a-\ to  the  form  of  a  fraction. 

n 


an-\-}p- — cd 
Ans.   ■ 


n 


A.7i^  I  5fl^ 
7.  Reduce  Ix ^ —  to  the  form  of  a  fraction. 

o 

.        56z — A.11^ — 5a 
Ans.   :x . 


FRACTIONS.  53 

8.  Keduce  15a -. to  tlie  form  of  a  fraction. 

4:771 

Am.   ; ■ — . 

4:771 

'J  Q ']Jl 

9.  Reduce  7a — b -. to  the  form  of  a  fraction. 

471 

2^an — 4b7i — *Je-{-7ri 

Am.   7 ■ — . 

4n 

10.  Change  llm — 4.7i-\-^ -— ^  to  the  form  of  a  fraction. 

°  6m — lii^ 

.        33??z- — 2277171^ — 12vin-\-'^7i^-\-a^ 
3m — 27i'^ 

11.  Reduce  ^x^-^-by^ —  ,,  to  the  form  of  a  fraction. 

JiX — ^y 

lQx^^l()xy''—24xY—^^y'—d^—a' 
^^-  2x-lf • 

Case  IV. 

119»  To  represent  a  fraction  in  the  form  of  a  whole  or  mixed 
quantity. 

Rule.  Divide  the  numerator  by  the  deTwmmator  for  the  inte- 
gral part,  and  write  the  remainder,  if  any,  over  the  denomiTiator 
for  the  fractional  part ;  annex  this  to  the  integral  part,  and  it 
will  represent  the  quantity  required. 

EXAMPLES. 

27 
1.  Change  -^  to  a  mixed  quantity. 


8 

27 


:27-i-8=3#.    Ans. 


2.  Chano;e  -—  to  a  whole  number. 

o     21 


j^=88^11=8.   Am. 


3.  Change    "       -  to  a  mixed  quantity. 

ax-]~a' 


■.ax-\-d^  -r-x=a-\ — .   A7is. 


X  '  x 

5# 


64  ALGEBRA, 


4.  Change  — - —  to  a  mixed  quantity. 

ab — a^     ~- ,.     ^  c^       . 

— - — ■=.ao — a-^o^=a — -.   Ans. 
b  0 

5.  Change  — to  its  equivalent  mixed  quantity. 


Ans.    (f- — axA^x^ 


a-\-x 

6.  Chancre  — ; —  to  a  whole  number.         Atis.  x~ — xii+il^, 

^ 1? 

7.  Change  ~  to  a  whole  number.        Ans.   x''~~\-xy-\-y^. 

X     y 

9 
Q^X' X 

8.  Find  a  mixed  quantity  equivalent  to . 

Ans.    x^ . 

a 

Case  V. 

120#  To  reduce  a  complex  fraction  to  a  simple  one. 

Rule.  If  the  numerator  or  denominator,  or  both,  be  ivhole  oi 
mixed  quantities,  reduce  them  to  improper  fractions.  Then 
multiply  the  denominator  of  the  lower  fraction  into  the  numerator 
of  the  upper  for  a  neio  numerator  and  the  denominator  of  the 
upper  fraction  into  the  numerator  of  the  lower  for  a  new 
denominator  ;  or,  invert  the  denominator  of  the  complex  fraction 
vjhen  reduced,  and  place  it  in  a  line  with  the  numerator ;  then 
tnultiply  the  two  numerators  together  for  a  new  numerator ,  and 
the  two  denominators  together  for  a  new  denomirmtor . 

All  fractions  in  this  proposition  must  be  reduced  to  this  form, 

a        3 

c        4 

-,  or  -,  before  they  can  be  solved  by  the  above  rule.     Now, 

b       5 

every  fraction  denotes  a  division  of  the  numerator  by  the 
denominator,  and  its  value  is  equal  to  the  quotient  obtained  by 
such  a  division.     Hence,  by  the  nature  of  division,  we  have, 


FRACTIONS.  55 


c      a     b      ah 

d     c     d     cd' 


Bj  the  preceding  rules  we  are  enabled  to  show  all  the  vari- 
ations that  can  possibly  happen  in  preparing  fractions,  and  also 
the  method  of  reducing  them  to  their  lowest  terms. 

EXAMPLES. 

7  7 

1.  Reduce  f  to  a  simple  fraction.       ^='^X.i=i^-     -A.ns. 

7-1 

2.  Keduce  ^  to  a  simple  fraction. 

71         29 
*¥ ~_£^ 

8J-       -1-7. 


Li. liT 59v    2    — 58 29  i^T^o 

^1 17 4'AtT  —  WQ — in"*       -^^t^' 


7 
3.  Reduce  -  to  a  simple  fraction. 


7     ^ 

1 1      1/^1 1   '"-'- 


Ans. 


8 

4.  Reduce  -^  to  a  simple  fraction. 

3.  3. 

5  5 


'¥       "4 

a 


g3 _2X 5^2  7 lTf5 ?f5'        -^'t^* 


5.  Reduce  — --  to  a  simple  fraction. 


b  b        a        \  a 


m-\-n     m-\-n     b     m-\-n     bm-\-bn 


6.  Reduce  to  a  simple  fraction. 

y 

a 


"   ^^     =%xA-  =  ^.    Am. 


a     xy-{-a     1     xy'\-a     xy-\-a 

^    y      y 


56  ALGEBRA. 


7.  Keduce to  a  simple  fraction. 

m 

X 

n 


,  h       ac-\-b 


c  c 


ac-4-b         n          acnA-bn       . 
= — !— X = ! — .     A71S. 


m     nx — m         c        nx — m     cnx — cm 
^ 

n         n 

8.  Reduce  ;^  to  a  simple  fraction.  Atis.     \^. 

X 

9.  Heduce  — ^r-  to  a  smiple  traction.     Ans.     „,  ,   .  . 

,  ,  %  ^  634-4?/ 

n 

^"~3  %m—n 

10.  Reduce  to  a  simple  fraction.  Aiis.  — ^ — 

2/— ^'+2 

11.  Reduce  — — to  a  simple  fraction.        „       o    i  ^ 

Case  VI. 

121.  To  reduce  fractions  to  a  common  denominator. 

Rule.  Multiply  each  numerator  into  all  the  denomiruitors 
except  its  oivn  for  a  new  numerator ^  and  all  the  denominators 
together  for  a  common  denominator. 

Or,  find  the  least  common  multiple  of  all  the  denominators, 
and  it  will  be  the  denominator  required.  Divide  the  common 
multiple  by  each  of  the  denominators,  and  multiply  the  quotients 
by  the  respective  numerators  of  the  fractions,  and  their  products 
will  be  the  numerators  required. 

IIRST   METHOD. 

1.  Reduce  yV?  i?  and  |,  to  a  common  denominator. 

5X   8x4=160,  numcHkr  for  tV=^|£. 

7X12X4=336,  numerator  for  I  =ff|-. 

3X12X8=288,  numerator  for  f  =§||. 

12x   8x4=384,  common  denominator. 


FRACTIONS. 


57 


Equimultiples  of  the  terms  of  a  fraction  express  the  same 
value  as  the  fraction  itself.  The  terms  of  -^^  are  each  multi- 
plied by  8  and  4.  Hence  4^f  £^  has  the  same  value  as  ^^.  The 
same  may  be  observed  of  |-  and  f . 

2.  Reduce  -,  -,  and  — ,  to  a  common  denominator. 
0   a  n 

«X^X^=^<^^i=numerator  of  -= —  . 

h     bdn 

cX^X^=^c?^==  numerator  of -=--7-. 

a     bdn 

^X^X^=^^w=uumerator  of-=^-^- 

71      bdn 

i'Kdy^n=hdn-=.QommoB.  denominator. 


SECOND   METHOD. 

3.  Keduce  |,  y^^-j  ^^^  h  to  a  common  denominator. 
4)8,  12,  4 
2,    3,  1;  4x2x3=24,  common  denominator. 
24 


12 

4 


3x7=21,  numerator  for  |-  =f^. 
2x5=10,  numerator  for  •r2-=^2¥* 
6x1=  6,  numerator  for  ^  =2^4. 


4.  Reduce  -7-,  — ,  and  7—,  to  a  common  denominator. 
4x   X  hx 

X)4:X,  x^,  '^x 

4)4,    X,   8 

1,    X,  2;  a: X^X'^X 2= 8:?;^  common  denominator. 

Sx"" 

lax 


Ax 


x' 


Sx 


2xX,(i=2.ax,  numerator  to  7-=^   ,. 

4x     Sx^ 

8  X^=83,  numerator  to  —=;::-—. 

x^     Sx^ 

\f1  1/7 '7* 

xyJoa=.^ax^  numerator  to  „   — -— r-. 


5.  Reduce  |-,  ^^^-j  ^"^^  h  to  a  common  denominator. 

Ans   A6    21    J.8 


68 


ALGEBRA, 


6.  KeducGy^y,  -j^g-,  ^,  and  7,  to  a  common  denominator. 

Ans. 

7.  Heduce  f  of  7|-  and  -^j-  of  5  to  a  common  denominator. 

Ans. 

8.  Reduce  f  of  -^j  of  17  and  J-  of  19  to  a  common  denom- 
inator. Ans. 

9.  Reduce  -  and  -^r  of  ;|-  to  a  common  denominator. 

5  "^  ^  7  1 


/I710         8  6  13  240_ 

yi/Z.6.     -rs-T-^n-j  TITBIT' 

mmon  denominato 
33^:^ — ocx^  4:bmy — 4icmy       axy 


10,  Reduce  — ,  — ,  and  •; ,  to  a  common  denominator. 

y     X  b — c 

Ans. 


11.  Reduce  — , --,  and  ,  to  a  common  denominator 

X  X — 2  y 

Ans. 


bxy — cxy  '      bxy — cxy    '  bxy — cxy 
d  ,  to  a  common  denominator. 

y 

axy — 2<2?/        bxy        doi? — Zo? — IdxAr^'^ 
x^y — ''Ixy  x^y — '^xy  x\j — ^Ixy 

12.  Reduce , -,  and     ^,,  ,  to  a  common  denominak)r 

X      y — 2  18 

18g?/+18%— 36^—363  h\x—\Ux  x''y—7xy—2x''-\-14:c 
ISxy — SQx  ^ISxy — 36:c'         ISxy — 36^; 


Ans. 


13.  Reduce  - — ^,  -,  -,  and ^,  to  a  common  denominator. 

b — 6   b   X  X — 0 

^abx^ — 20(z3z  abx^ — Zay? — habx-\-\hax 


Ans.  • 


b''x'—3bx^—bb~x-{-15>bx  3  V— Sbx'—bb^x+lbbx 

b''dx—Udx—Wd-\-\bbd  ab''x—b^x—^abx-\-Wx 


.  Wx'—Ux'—bb'x-^lbbx  b'x'—dbx'—bb''x-]-lbbx 

1  -,  to  a  common  denominator. 

2/— 3 

x^y — 3^;^  xy"^ — "^xy  ay — oa        ax 

xy — Zx  '   xy — 3a;  '  xy — 3a;'  xy — 3a;* 


14.  Reduce  x,  y,  -,  and  -^  to  a  common  denominator. 

X  y—3 

Atis. 


15.  Reduce  a,  b,  c,  d,  and  -,  to  a  common  denominator. 

o 

ab   b"   be   bd  a 

'^''  Tl'l'  T'  ~b' 


FRACTIONS.  59 


X 


y          3?/ 
16.  Chano^e  -  and  -:p  to  a  common  denominator. 

771  1 

2       m—n  .  6a;         . 

Atis,  -k —  and 


Smy  'S?ny 


X         5J- 
17.  Cliano;e  —  and  —  to  a  common  denominator. 
7^  X 


"&■■ 


2x~      ,  80 
Ans.     — r-  and  —;r-. 

lOZ  iDX 


Case  VII. 

ADDITION    OF    FRACTIONS. 

122.  To  add  fractional  quantities. 

KuLE.     Reduce  the  fractioris  to  a  common  denominator^  and 
write  the  sum  of  the  numerators  over  the  common  denominator. 

EXAMPLES. 

1.  Add  1^,  y^j,  and  J-^,  together. 
[lere    7x12x16=1344  ^ 

^X   8X16=  640  >  the  new  numerators. 
llX  8x12=1056) 


3040 


:1|^.     Ans. 


And     8x12x16  =  1536,  the  common  denominator. 

2.  What  is  the  sum  of  -,  -,  and  -  ? 

b    d  f 

Here  aXdXf=cidf  \ 

cX^X/=c^/"  /  *"G  new  numerators. 

eX'^Xd=ehd  ) 
And    by^dycf^=bdf,  the  common  denominator. 

™        „        adf  .  cbf  ,  ebd      adf-\-cbf-\-ebd 


60  ALGEBRA. 

...     .  .  .  Sx'      ,  ,    ,2ax 

3.  Add  tlie  following  quantities,  a 7-  and  b  -\ . 

0  c 

Sx""     ab—Zx^     ,  ,  lax     hc-\-'lax 
b  b  c  c 


ah — Zx^yc^c^iahc — ocx~   ) 

,0    .  ^  ,     (  numerators. 

bc-Ylaxy^b^b'c^-^abx  ) 

by^c  =:bc^  common  denominator. 

abc—2>cx'     b\-[-2abx     abc—Zcx'-\-Wc^1abx        ,  _  , 

be  be  be 

2ahx — Zcx^        . 

^ .     Ans. 

be 

4.  Add  together  g^,  ^,  and  ^, 

84gV+8Q^7?z?z"4-105a^^e 
^^'  U^adn^ 

5.  What  is  the  sum  of  ^,  -jV,  and  |-?  Aois.  Iff. 

6.  What  is  the  sum  of  f,  y\,  and  f  ?  Ans.  2^2_7^. 

7.  What  is  the  sum  off,  f,  |,  and  -J^l  Ans.  2^|. 

8.  What  is  the  sum  of  8f ,  3f ,  and  7|  ?  Ans.  20i. 

9.  What  is  the  sum  of  |  of  7^,  and  -/_  of  13  ?   Ans.  13/^. 

10.  What  is  the  sum  of  f  of  1,  and  J-  of  f  ?  Ans.  -fi. 

11.  What  is  the  sum  of  |  and  -|-  ?  Ans.  f|. 

12.  What  is  the  sum  of  f  of  ^  and  f  of  ^  ?     Ans.    if  §9. 

.«    -r^.    i    ,  ^3:r       ,  2:c  .         ^exA-^ax 

13.  Find  the  sum  of -j—  and  ^r— .  ^?w. 


4fi^  3e  *  '        Viae 

XXX  ^        47a: 

60" 

..    T..   -.   1  ^4a      ,  ft-3  .        23^^-21 

15.  Find  the  sum  01  -     and  — ; — .  A7is. 


14.  Find  the  sum  of  k,  -7,  -.  Ans.   ^^. 

o   4   0 


7  4    •  28 


lo.  Find  the  sum  of  47?z,  — r: — ,  and 


3 

9^  +  24772  +  871  +  1 


Ans. 

o 


FRACTIONS.  61 


17.  What  IS  the  sum  oi  — ^ — ,  — - — ,  and  —  ? 

0  o  A 

139a-8 
Am.   — ^TT — . 
3  3  ^^ 

18.  Add  — -—  and tocrether.  a^ 

a^h         a-h  ^^        ^« 


a      a — b  a-\-b 

— -,  — — ,    and 

a—o  c-\-d  c — a 

2a'c-\-ac'—ad'-2ahc-{-2abd—2b'd 


19.  Add   ■ — -,  — ^— „    and  — -^^  together. 


ac-—ad^-\-bd^—bc^ 

o/^    All     3^-  c  ,       4ac— 2^c+9a5^— 93^ 

20.  Add to -.  A71S.  -jr^-, Q  ,3    ,  QZ4    • 

a—b       2a— b  babe — \)abc-\-6bx 

~2~      ""3" 

Case  VIII. 

SUBTRACTION    OF    FRACTIONS. 

128.  To  subtract  one  fraction  from  another. 

Rule.  Reduce  the  fractions  to  a  commoii  denominator,  sub- 
tract the  numerator  of  tlie  subtrahend  from  the  numerator  of  the 
minuend^  and  write  the  diffei-ence  over  the  common  denomi- 
nator. 

EXAMPLES. 


1.  From  I  take  -/y. 
=77) 
4X   9=36  )  ^' 
And  9x11  =  99,  the  common  denominator. 


Here  7x11=77  . 

'le  new  numerators. 


Whence  U—U=U'     ^^^' 


a     1      c 

2.  From  -  take  -. 

b  d 


Here  ay^d=-ad  . 

C  the  new  numerators. 


Xd=^ad  ) 
Xb=:bc    ) 
And  bXd^bd,  the  common  denominator. 


.^_  ad      be      ad — be       . 

Whence  r-.—  73= — 73 — .     Arcs, 
bd      bd         hd 


62  ALGEBRA. 

3.  From  |  take  j\.  Ans.  |f . 

4.  From  f  take  |.  Am.  f. 

5.  From  7f  take  4|-.  A?is.  S^. 

6.  From  Q^  take  |  of  5.  Ans.  2§^. 

7.  From  8|  take  f  of  17^.  ^?w.  1}. 

8.  From  f  of  ll^-^  take  -iJ-  of  3|.  A?is.  If  Jg-. 

9.  From  |  of  13f  take  /^  of  7^.  Ans.  9^\. 

10.  From  f  of  7  take  ^  of  17^.  Ans.  2^^. 

11.  From  ^  take  i.  ^7^..    ^^2_7_. 

12.  Take  — ;-  from  -A..  ^tw.       ^ 


fi^-j-l  fi — 1*  '   «^ — 1* 

13.  From  — -  take  — -.  -  Ans.    — — . 

5  7  35 

^,    ^        3a— 23     ,     2a— U 

14.  irom  — rr —  take 


3c  5^. 


l^ab—Qac—10b''+12bc 
Ibbc.  • 


15.  Required  the  difference  of  -^  and  — .        A?is.     -— -. 

16.  Subtract  ^  from  ^±^.  ^72^.       ^^^ 


a;-j-2/ ""  '2; — ?/'  '     a;"^ — y^' 

-ihT    c(  1                   ^ — ^  n         o       fl^+3  ,           ^       2a 

17.  Dubtract  a —  irom  da-] r— .  ^7W.     la-\---. 

ad  a 

18.  Subtract  z —  from  Ix ^ — .  Ans.  (Sx-{-a-\-j;. 

Z                              a  b 

19.  From  ^    '    '  take  ^^ — -^.  Am.     4. 

g;-!-^  « — ^ 

"^     .     ~X                     .  hd'-\-2^ah-k'U^ 

20.  From take  -.  Am.     —^ i —  . 

a — h          a-\-h  ba'^—bb^ 


FRACTIONS.  63 


Case  IX. 

MULTIPLICATION   OF    FRACTIONS. 

121 1  To  multiply  fractions  together. 

Rule.  Multiply  the  numerators  together  for  a  new  numer- 
ator^ and  the  denominators  for  a  new  denomi7iator. 

Whe?i  the  numerator  of  one  of  the  fractions  and  the  denom- 
inator  of  the  other  can  he  divided  by  some  quantity  which  is 
common  to  each  of  them^  the  quotients  may  he  used  instead  of  the 
fractions  themselves. 

Also,  ivhen  a  fraction  is  to  he  multiplied  hy  an  integer,  it  is 
the  same  whether  the  numerator  is  multiplied  hy  it  or  the  denom- 
inator is  divided  hy  it. 

If  an  Integer  is  to  he  multiplied  hy  a  fraction,  or  a  fraction  hy 
an  integer i  the  integer  may  he  considered  as  having  unity  for  its 
denominator. 

A  mixed  quantity  should  he  reduced  to  an  improper  fraction. 

Powers  or  roots  of  the  same  quantity  are  multiplied  together 
hy  adding  their  indices. 

EXAMPLES. 

1.  Multiply  f  by  I.  iXl=U-    Ans. 

^    ^^  ^  .  .    a  .     m  a     m     am        . 

2.  Multiply  -  by  — .  _X-=7— •     -^^s. 

^  ''  h        n  h      n      hn 

^    ,r  1  .  1    cibc  ,     mh  ahc     mh     ah       . 

3.  Multiply  —  by—,.  —  X7-i=— ,-    Ans. 

mn       hcd  mn     oca     tia 

Note.  —  The  Z>,  c  and  m,  are  cancelled  in  both  factors. 

^    ,^  ,  .  ,      3a  ,      ^  3a      2m     ^am       . 

4.  Multiply  — -  by  zm.  =^X-^i-=;;T--     ^1^^^- 

^  ^   Ihc    -^  7hc       1        The 

5.  Multiply  a+—  by  -. 

hy     a^-j-hy     a^-\-hy     m     c^m-\-hmy        . 

a-\ — ^= *  X — -^ •     -ii-ns. 

a  a  a         n  an 


64  ALGEBRA. 

6.  Multiply  ^^^  by  ^-^^. 

^  ^  x+y     ^  x+y 

m-\-7i     77i-\-n     m'^-\-2mn-\-Tt^ 

7.  Multiply  ^:  by  4^.  Ans.    ^. 

mm        innd  mttd 

8.  Multiply  ^  by  g^.  .   ^,«.     ^^. 
y.  Multiply  —-  by  ^rj-r.,  Ans.     ^^^  ,„  ,. 


- —  by  — -o.  Ans.      -j-, 

hy  ay^  ahy^ 


11.  Multiply  5^  by  g^,.  An..     1. 

When  the  multiplier  and  denominator  of  the  fraction  are  the 
same  quantity,  they  cancel  each  other. 

12.  Multiply  -^  by  "Imn. 

13.  Multiply  ^  by  lla3. 

-I  A       T./r    1    .     1       4(2CC?  , 

14.  Multiply  — —  by  xy. 

xy 

15.  Multiply  —^  by  17  ab. 

16.  Multiply  ^^ab  by  --;^-. 

17.  Multiply  -^^  by  —  . 

irni^        mri^ 

18.  Multiply';^"  by  il^. 

^  ^  ihy     ^  m-rt 

19.  Multiply-   by—.  Am.     —r^. 

^  ''  b      -^   h  bh 


Ans, 

,     Sab. 

Ans. 

Qmn. 

Ans. 

4acd. 

Am. 

2)hm. 

Am.     X. 

5 

ab^ 

mhi^ 

A?is.     ^. 

Am. 

^m+n 

FRACTIONS. 


65 


Case  X. 

DIVISION    OF   FRACTIONS. 

125.  To  divide  one  fraction  by  another. 

Rule.  Multiply  the  denominator  of  the  divisor  by  the  nu?ner- 
ator  of  the  dividend  for  the  numerator,  and  the  numerator  of  the 
divisor  by  the  deiwminator  of  the  dividend  for  the  denominator. 

Or,  invert  the  divisor,  and  proceed  as  in  multiplication. 

Or,  divide  the  numerators  by  each  other,  and  the  denominators 
by  each  other,  when  this  can  be  done  without  a  remainder. 

Mixed  quantities  should  be  changed  to  improper  fractions. 

EXAMPLES. 

-    -r, .  . ,    « ,     3a  <z     3a     a      8       8a      2       . 

1.  Dmdejby-^.  -^_=-Xg^=j2^=3.    An,. 


2.  Divide  --  by  -r^. 

2b    ''  id 


3a     5c      3a     4d     12ad     Qad       ^ 
2b  •  Ad'' 2b  ^  5c  ""lOic  ~bbc ' 

3.  Divide  1^^  by  ^^*. 

3a— 2^    -^    4:a-{^b 

2a-f3      4a-^b  _8a^-}-6a^4-3^ 
3a— 2^^3a+25~    Qa^-ibl    ' 

4.  Dmde-by-.  Ans.     — =-. 

5.  Divide  -^  by  Bx\  Am. 


5      •"        ••  15a;'^""5* 

6.  Divide  -— -  by  -— .  Ans. 


2      '   13  ^^'^'      6n'' 

7.  Divide  1^  by  11.  Ans.     ^. 

6       ''  6 

8.  Divide  —  by  xy.  Am.     — . 

^y  x'Y 

9.  Divide -^  by  ^  Ans.     ~^. 

6^ 


66  ALGEBRA. 

10.  Divide  -^  by  y.  Ans.     -|. 

-.-.     T..  .-,       4:r     ,        2  ,  2^^+2:c 

11.  D,,,de  ^^-^  by  ^.  ^^.     -^—^. 

12.  Divide  —. by  th-^-       Ans. ^  ,  ^ . 

13.  Divide  ^^.__^^^^^3.  by  -^^-^. 

20a^— 20c^^-20a^y-W 
12a^— 14a^<^— 8a'^^»-^+10a3^* 


-4?w. 


NEGATIVE   EXPONENTS. 

If  we  divide  a^  successively  by  a,  the  following  will  be  the 
quotients  : 

1111 

a\  f^^  a^,  a\  1,  -,  — ,  -,  -,  &c. 
a  a^  a    a^ 

By  examining  the  above,  we  perceive  that  the  exponent  of 
each  term  is  one  less  than  the  preceding ;  therefore  the  division 
might  have  been  expressed  thus : 

a^,  a^,  a^,  a^,  a",  a~^,  a~^,  a~^,  ar^. 

By  comparing  the  last  quotients  with  the  former,  we  find, 

a  a^ 

1  1 

We  also  perceive  that  exponential  quantities  are  divided  by 
subtracting  their  indices. 

Hence,  if  ar^  be  divided  by  a~^,  the  quotient  will  be  ar^~^z= 
or^ ;  or,  rc"'"  by  :r-"=a;-'"-". 

We  also  infer  from  the  above  illustration  that 

«•'   a}   c^       _,     _2     _3     1   1    1 

(T  a/  a/  a  a^  w" 

Again,  we  see  from  the  above  that  any  quantity  which  has 
zero  for  its  exponent  is  equal  to  1. 


FRACTIONS.  67 

We  infer,  also,  that  if  similar  quantities  with  negative  ex- 
ponents are  divided  by  subtracting  their  indices,  that  such 
quantities  are  multiplied  by  adding  their  indices. 

Thus,  a-''Xar^=ar\  and  a^Xcir^=a^z=a'=l. 


EXAMPLES. 

1. 

Divide  a~^  by  fi^~^. 

Ans.   a~'\ 

2. 

Divide  mr^hj  mr^^^ 

A71S.  m^. 

3. 

Divide  x^  by  x~^. 

Ans.  7?. 

4. 

Divide  lx~^  by  x~^. 

Ans.  7x. 

5. 

Divide  8?/-^  by  'f. 

Atis.  8?/-^. 

6. 

Multiply  a-'  by  7a-^ 

Atis.  7a~^. 

7. 

Multiply  om  by  m~^. 

Atis.  Bm~^. 

8. 

Multiply  4cx-^  by  x\ 

Atis.  4:X~\ 

9. 

Multiply  a-\b-''.  c-^yy  a\ 

bU\ 

Atis.  a^  b  c~\ 

10. 

Divide  a^  hj  a  ^. 

t 

3 1 
Atis.     a^^. 

11. 

3.               —4 

Multiply  n^  \>Y  71  ^. 

Alls.        7^~'^. 

To  free  fractions  from  negative  exponents. 

Rule.  TraTisfer  the  letters  which  Jmve  Tiegative  exponents  in 
the  numerator  to  the  denominator^  and  those  vjhich  have  Tiegative 
exponents  in  the  denominator  to  the  numerator,  and  then  change 
the  sign  of  the  exponent. 

Note.  This  rule  implies  the  multiplying  of  all  the  terms  of  the  numer- 
ator and  denominator  by  the  same  quantity.  Therefore,  by  Art.  121,  the 
value  of  the  fraction  is  the  same. 

EXAMPLES. 

1.  Free  the  fraction  _^  from  negative  exponents. 

A        d~e 
Ans.  -^73. 

1-3  /v,-2 


m  n —  7) 
2.  Free  the  fraction  —       _^  from  negative  exponents. 

X  "Ij      z 


Ans. 


my^z 

XTV'jf 


68  ALGEBRA. 

3.  Free  the  fraction  — ^  _^  from  negative  exponents. 

Tit  iL      C 

Am.   ^—f- — . 

4.  Free  the  fraction _.^  _^^  _^  from  negative  exponents. 


Ans. 


93''  ^     1  9^4 


^      O      I 

5.  Free  the  fraction  — ^- from  negative  exponents. 

Ans. 


6.  Free  the  fraction    ,1,3  _^,_^  3  from  negative  exponents. 

We'd' 
Am.   — 2^-. 


SECTION    VII. 

EQUATIONS. 

Art.  126.  The  doctrine  of  equations  is  that  branch  of 
Algebra  which  treats  of  the  method  of  determining  the  values 
of  unknown  quantities  by  means  of  their  relations  to  others  that 
are  known. 

This  is  effected  by  making  certain  algebraic  expressions 
equal  to  each  other ;  which  formula,  in  that  case,  is  called  an 
equation. 

127.  The  terms  of  an  equation  are  the  quantities  of  which 
it  is  composed ;  and  the  parts  that  stand  on  each  side  of  the 
sign  =  are  called  the  two  members,  or  sides,  of  the  equation. 

Thus,  if  x=LaA^h^  the  terms  are  x^  «,  and  h  ;  and  the  mean- 
ing of  the  expression  is,  that  some  quantity  .t,  standing  on  the 
left  side  of  the  equation,  is  equal  to  the  sum  of  the  quantities 
a  and  3,  on  the  right  side. 


EQUATIONS.  69 

128.  A  simple  equation  is  one  which  contains  only  the  first 
power  of  the  unknown  quantity  ;  as, 

X 

rc-|-<z=10;  c2;-[-32:=c  ;  or,  42;-f-=17  ; 

in  which  equation  x  denotes  the  unknown  quantity,   and   the 
other  letters  and  the  numbers  the  known  quantities. 

129*  A  compound  equation  is  one  which  contains  two  or  more 
different  powers  of  the  unknown  quantity ;  as,  2;--f-(22'=c? ;  or, 

130,  A  quadratic  equation  is  one  in  which  the  highest  power 
of  the  unknown  quantity  is  a  square. 

131 9  A  cubic  equation  is  one  in  which  the  highest  power  of 
the  unknown  quantity  is  a  cube  ;  as, 

a;^=64  ;  or,  x' — axr-\-hxz=c. 

182,  The  root  of  an  equation  is  such  a  quantity  as,  being  sub- 
stituted for  the  unknown  quantity,  will  make  both  sides  of  the 
equation  vanish,  or  become  equal  to  each  other. 

133.  A  simple  equation  can  have  but  one  root;  but  every 
compound  equation  has  as  many  roots  as  it  has  powers. 

134.  Identical  equations  are  those  which  have  the  terms  of 
the  equation  the  same. 

135.  Nu7nerical  equations  are  those  which  contain  numbers 
only  in  connection  with  the  unknown  quantities ;  as, 

a;2_i_7;^.^5=,100. 

136.  Literal  equations  are  those  in  which  numbers  are  repre- 
sented by  letters ;  thus, 

7?-\-px-\-ap=^r. 

18T,  To  reduce  an  equation  is  to  discover  the  value  of  the 
unknown  quantity  in  it. 

138.  The  process  of  reducing  equations  depends  upon  the 
following  simple  principles  or  axioms ; 


70  ALGEBRA. 

1.  If  to  equal  quantities  we  add  the  same,  or  equal  quantities, 
tlie  sums  will  be  equal. 

2.  If  from  equal  quantities  we  subtract  the  same,  or  equal 
quantities,  the  remainders  will  be  equal. 

3.  If  we  multiply  equal  quantities  by  the  same  quantity,  the 
products  will  be  equal. 

4.  If  we  divide  equal  quantities  by  the  same  quantity,  the 
quotients  will  be  equal. 

5.  If  we  extract  the  same  roots  of  equal  quantities,  those 
roots  will  be  equal. 

6.  If  we  raise  equal  quantities  to  the  same  powers,  those 
powers  will  be  equal. 

139 1  The  known  and  unknown  terms  of  an  equation  may  be 
combined  in  various  ways. 

1.  By  addition ;  as,  x-\-7=lQ,  or  x-{-a=b. 

2.  By  subtraction;  as,  x — 9=19,  or  x — ^=5. 

3.  By  multiplication;  as,  7:c=84,  or  ax=c. 

X  X 

4.  By  division ;  as,  -=12,  or  -=d. 

-^  4  a 

5.  By  a   combination  of  two  or  more   of  these  rules ;    as, 

ijX    ,    ^  1—         f,  »■  tiX 

— -j-l7=:o:r— 5;  or, ---\-77i=cx—7i. 

I. 

140.  To  find  the  value  of  the  unknown  quantity,  when  com- 
bined with  a  known  quantity,  by  addition  or  subtraction. 

1.  Let  :r-|-7=:16 ;  and  it  is  required  to  find  the  value  of  x. 

Now,  as  x-\-7  is  equal  to  16,  it  is  evident,  from  the  second 
axiom,  that,  if  from  each  of  these  equal  quantities  we  subtract 
the  same  quantity,  the  two  remainders  will  be  equal.  We 
therefore  subtract  7  from  each  member  of  the  equation. 

Thus,  :c+7— 7=16— 7. 

As  the  plus  7  and  minus  7  in  the  first  member  of  the  equa- 
tion cancel  each  other,  the  equation  will  be 

a;=16-7=9. 


EQUATIONS.  71 

Therefore  the  value  of  x  is  9 ;  but,  in  the  operation,  we  have 
only  transposed  the  plus  7  from  the  first  member  of  the  equation 
to  the  second,  and  changed  it  to  a  minus. 

2.  Again,  let  :c — 5=12;  it  is  required  to  find  the  value 
of  X. 

Now,  by  the  first  axiom,  we  find,  if  equals  be  added  to  equals, 
their  sums  will  be  equal ;  we  therefore  add  5  to  each  member 
of  the  equation,  and  we  have 

:^;_5_{_5=12-|-5. 
In  the  first  member  of  the  equation,  we  have  — 5  and  -}-5  ;  and, 
as  they  will  cancel  each  other,  the  equation  will  stand 

a;=12+5=:17. 

Therefore,  the  value  of  x  is  17. 

All  that  we  virtually  have  done  in  the  above  operation  has 
been  to  transpose  the  minus  5,  in  the  first  member  of  the  equa- 
tion, to  the  second,  and  to  change  it  to  plus. 

From  the  foregoing  examples  and  illustrations,  we  deduce 
the  following 

Rule.  When  a  quantity  is  trayisposed  from  one  memher  of 
the  equation  to  the  other ^  the  signs  must  he  changed. 

3.  Given  a;-f-15— 5=86— 8  to  find  the  value  ofx. 
By  transposing,       a: =86 — 8 — 15-}-5. 

By  uniting,  2:=68. 

4.  Given  a;— 29+3=100— 19+3  to  find  the  value  ofx. 
By  transposing,        2:=100-19+3+29--3. 

By  uniting,  :c=110. 

5.  Given  a;+12— 3=7 — 4  to  find  the  value  of  x. 
By  transposing,        a;=7 — 4 — 12+3. 

By  uniting,  x=^ — 6. 

6.  Given  a:— 5— 4^24+7  to  find  the  value  of  z. 

Ans.  a:=40. 
II. 

Ill,  When  the  known  and  unknown  quantities  are  combined 
by  multiplication. 


72  ALGEBRA. 

7.  What  is  the  value  of  a;  in  the  equation  bx-\-lS=bSl 
By  transposition,         bx=5S — 18. 

By  reduction,  5a:=40. 

By  division,  r=:8. 

We  say  that  if  5  times  x  is  equal  to  40,  it  is  evident  that  -| 
of  5  times  x,  that  is,  x,  is  equal  to  8. 

142.  Hence,  if  the  unknown  quantity  in  any  equation  be 
multiplied  by  any  number  or  quantity,  in  order  to  find  its  value, 
we  divide  the  sum  of  all  the  quantities^  after  being  reduced,  by 
the  coejficient  of  the  unknown  quantity. 

8.  What  is  the  value  of  x  in  the  following  equation, 

7.r— 28==46-f  10  ? 

By  transposition,         72-=464-10-|-28. 
By  reduction,  7:r=84. 

By  division,  a:=12. 

9.  Given  4a;— 5=71+8  to  find  x. 

10.  Given  Gz— 17— 7=0  to  find  x. 

11.  Given  5^:4-28+8=6  to  find  x. 

12.  Given  7:r-17+3=100  to  find  x. 

13.  Given  23a;— 96+1=0  to  find  x. 

14.  Given  17a;— 7 — 5 — 8=4  to  find  x. 

15.  Given  9a;=7+8+10  to  find  x. 

16.  Given  7a;— 10=52;+14  to  find  x.  Ans.  12. 

III. 

113.  To  reduce  an  equation  when  the  known  and  unknown 
quantities  are  combined  by  division. 

X 

17.  Given  t=8  to  find  the  value  of  a;. 

4 

Multiplying  both  terms  by  4,  we  have  a;=32. 
Therefore,  if  both  terms  of  an  equation  be  multiplied  by  any 
number,  their  products,  by  axiom  third,  are  equal. 


Am 

.  21. 

An^.  4. 

Atis. 

-6. 

Ans. 

16f. 

Atis. 

4/^. 

Ans. 

ItV- 

Ans 

^• 

EQUATIONS.  73 

141.  If  a  fraction  be  multiplied  by  its  denominator,  the 
product  is  the  numerator,  and  the  denominator  disappears. 

3a; 

18.  Given  -;^=9  to  find  the  value  of  5;. 

0 

Multiplying  by  5,  8:r=45. 

Dividing  by  3,  3;=15 

19.  Given  ~t=c  to  find  the  value  of  x. 

a 

Multiplying  by  d^  ax=:.cd. 

cd 

Dividing  by  «,  x=—. 

20.  Given  -4-7r — -r-=17  to  find  the  value  of  a;. 

2      o       5 

A.X    62; 
Multiplying  by  2,  a;-}--^— y=34. 

18a; 
Multiplying  by  3,  3a;+4z ^-=102. 

Multiplying  by  5,  15a;+20a;-18z=510. 

Uniting  the  terms,  17a;=510. 

Dividing  by  17,  a:=30. 

145.  Hence  an  equation  may  be  cleared  of  fractions  by 
multiplying  each  term  of  the  equation  by  the  several  denom- 
inators. 

21.  Given  ^+-^ ^, — ^^=^  to  find  the  value  of  2:. 

4   '   6       8      12 

The  least  common  multiple  of  the  denominators  4,  6,  8,  and 
12,  is  24;  and,  multiplying  each  member  of  the  equation  by 
this  number,  we  obtain  ^ 

18a;-}-20a;— 92'— 2a;=216. 
Uniting  the  terms,  27.^=216. 

Dividing  by  27,  a;=8. 

146.  Hence  an  equation  may  be  cleared  of  fractions  by  mul- 
tiplying each  term  of  the  equation  by  the  least  common  multiple 
of  the  denominators. 

7 


74  ALGEBRA. 

22.  A  boy  being  asked  Low  many  cents  he  had,  replied,  that 
if  he  had  f  and  ^  as  many,  in  addition  to  what  he  now  had,  he 
should  have  62.     Required  the  number  he  had. 

Let  X  represent  the  number. 

Then,  ^+5|+:,=62. 

By  multiplying  all  the  terms  of  the  equation  by  the  least 
common  multiple  of  the  denominators,  4  and  6,  which  is  12,  we 
have 

9xi-10xJ[-12x=7U. 
Collecting  the  x's,  31a;=744. 

Dividing  by  31,  x=  24.     Ans. 

VERIFICATION. 

£424     5X24     ^^^g2_ 
4  b 

18+20  +  24=62. 

23.  Given  l^Z^+3=6  to  find  x. 

4      ' 

Multiplying  by  4,  15— 2:+12=:24. 

Transposing,  15+12 — 24==x. 

Changing  terms,  :c=15+12 — 24. 

Reducing,  ^=3. 

24.  Given  -— ^7-=^^  *o  ^^^  ^' 

3^. 9 

Multiplying  by  3,  5a;— 4 ^-^^—=39. 

Multiplying  by  2,  lOx— 8— 3.7;+9=:78. 

Transposing,  10.r— 3.2:=78+  8—9. 

Collecting  terms,  7a:=77. 

Dividing  by  7,  a:=ll.      A?is. 

_^     „.        7JIX — n     -        _    , 

2o.  Given =6  to  fina  x. 

a 

Multiplying  by  a,  mx — 7i=ab. 

Transposing,  mx=:ab-{-n. 

Dividing  by  m,  x= — ^^.     Aiis. 


EQUATIONS.  75 

IV. 

147.  Combining  the  foregoing  rules  and  illustrations,  we 
deduce  the  following 

General  Kule  for  solving  all  Simple  Equations  which  contain 
only  one  unknown  term  : 

1.  Clear  the  equation  of  fractions. 

2.  Transpose  all  the  terms  containing  the  unknown  quantity 
to  one  side  of  the  equation,  and  all  the  remaining  terms  to  the 
other  side,  and  reduce  each  member  to  its  most  simple  form. 

3.  Divide  each  member  of  the  equation  by  the  coefficient  of  the 
unknofwn  term, 

19     6x 

26.  Given  2^- — 7-=-j — 1-4  to  find  x. 

4       4 

Multiplying  by  4,  8:^—19=3.^+16. 

Transposing,  8a; — 3a:=16-|-19. 
Collecting,  5a:=35. 

Dividing  by  5,  x=:  7. 

27.  Given = — - —  to  find  x. 

a  0 

d (t^X 

Multiplying  by  a,  1 — bx  •= — - — . 

Multiplying  by  3,  b — b^x=a — c^x. 

Transposing,  a^x — b'^x=.a — h. 

Dividing  by  a^— 52  x=— — -=— — . 

a^ — b^     a-\-b 

28.  Given  ^J^^-^'^^h--  to  find  x, 

bx     dx      bdx  X 

Multiplying  by  bdx,  ad-\-bc — [a—c)=bdhx — bd. 

Omitting  the  parenthesis,       ad-{-bc — a-\-c=zbdhx — bd. 
Changing  and  transposing,  bdhx=ad-{-bc-\-bd — a-\-c. 

ad-\-bc-\-bd — a-[-c 


Dividing  by  bdh^ 


bdh 


76  ALGEBRA. 


V. 


118.  If  the  terms  of  the  equation  contain  both  simple  and 
compound  denominators,  it  will,  generally,  be  found  convenient 
to  divest  it  of  the  simple  denominators  at  first,  and  afterwards 
of  those  which  are  compound. 

^_     „.        6a:+7  ,  72:4-13     1x-\-^       ^    , 

29.  Given  _^+^=_|l_  to  find  x. 

Multiplying  by  9,  62' +7 +^^^^=62: 4-12. 

K>X-\-o 

Qox-\-Wl 
Transposing,  ^     .  ,,    =6.r-j-12— 6a;— 7=5. 

bx-{-6  ' 

Multiplying  by  6a:-|-3,  63z4-117=30a;-f-15. 

Transposing,  63a;— 30a;=15— 117. 

Keducing,  33a;=— 102. 

D  ividing,  x=. — oj\. 

_.    ^.        2ar4-8^      13a;-2      x     7x    x+lQ  ^    _    . 

30.  Given  —^-^,^^-^-=-—^  to  find  x. 

Multiplying  all  the  terms  by  36,  it  being  the  least  common 
multiple  of  9,  3,  12,  and  36,  we  have 

Sx^M-  :,^  i  12a;=21a;-a;-16. 

J.  IX  —  O-j 

-r.   ,     .  rn     468^;— 72 

Keducing  terms,  oU=— — — -. 

17a; — o2 

Multiplying  by  17.2;— 32,       850.r-1600=468a;— 72. 

Reducing  terms,  382a:=:1528. 

Dividing  by  382,  a:=4.     Ans. 

EXAMPLES. 

1.  Given  5a;-}-22— 2a;=31  to  find  x.  Ans.  a;=3. 

.2.  Given  4— 19a;=14— 21a;  to  find  x,  Ans.  a:=5. 

3.  Given  24a;— 12=240— 12a;  to  find  x.  Ans.  a==7. 

4.  Givenl5a;-|-7a;— 10=12a;-}-90tofinda;.  Ans.  a:=10. 

5.  Given  7a;-l-2a;=12a;— 36  to  find  x.  Ans.  a;=12. 


EQUATIONS.  77 

6.  Given  12a:— 3a;— 2:?;=63  to  find  x.  Am.     a;=9. 

7.  Given  a:-|-r+^=87  to  find  x.  Am.     a:=60. 

'  4  '  5 


108. 


X  X 

8.  Given  a:— -r+13=;^-["40  to  find  x.        Am.     x= 

9.  Given  ^-{-^=:^-|-22  to  find  x.  A7is.     a:=120. 

10.  Given  a;--+20=^4-^+26  to  find  x.    Am.     x=b6. 

7  2  '  4  ' 

11.  Given  3a:+^4-15=|4-41  to  find  x.        Aiis.     a;=8. 

4  2 

12.  Given  a; ^^^-=8  to  find  x.  Am.     a;=28. 

b 

3a:— 11     5a:— 5     97— 7a: 

13.  Given  21-j — — = — j —  to  find  x. 

JLO  o  Ji 

Alls,     a: =9. 

3^ 5  2a: 4 

14.  Given  x-\ — =12 — ^^^—- —  to  find  x.    Atis.     a:=5. 

2  o 

15.  Given  n:c-^-^-^-^=.20x-^-o  to  find  x. 

tJ  0  2 

J.72^.     a::^2. 

16.  Given  9a: ^ — I ^ — =12a: ; 13  to  find  x. 

2*3  4 

Am.     xz=7. 

17.  Given  a:-f  ^+^-t-^+^=2a:4-17  to  find  x. 

jU        O        4:        0 

Alls,     a: =60. 

18.  Given  -=ih-{-c  to  find  x.  Arts,     x 


X  '  '  b-\-c' 

19.  Given  8a:— 40=0  to  find  x.  A?is.     x=b. 

20.  Given  a-\ — =b-\-c-^-  to  find  x.      Aiis.    x=. 


X  X  a — h — c 

„^     „.              3a:-3  ,   .     20-a:     6.a:-8     4a:-4  ^    „    .    . 
21.  Given  x }-4= — ^ = 1 ■= —  to  find  the 

value  of  a:.  Arts.     2'=:  6. 

7# 


78  ALGEBRA. 

22.  Given  ax^-\-hx=7nx^-\-nx  to  find  x.       Ans.  x=: 


a — m 


23.  Given  aa;+7?z=3a:-4-%  to  find  a:.  Atis.  x= . 

a — b 

ox     X  hciTTi c) 

24.  Given =m — c  to  find  x.         Ans.  x=—^^ . 

b      c  oc — b 

25.  Given =i\hx-\-n  to  find  x.  Ans.  x=z- ^-- — . 

a      m  Im — Ibam 

26.  Given  — ; =a — b  to  find  x. 

b  c 

.  4ab — ac-l-abc — Pc 

Ans.  xz= — 


b—c 


CiX  d 

27.  Given \-de=Sx to  find  x. 

b—c  e 


Atis.   x= 


cde^ — bde^ — bd-{-cd 

A71S.  x= — - — —T^ — ^7 . 

oce-\-ae — obe 

' —  to  find  a 

c 

2bcm-\-2bc?i — 4ab — 2ac — abc 


4x—a     2x-{-2a  2x-\-2a 

zo.  Given  ox ; i z=?n-\-n ' —  to  find  x. 

3       '        4  '  c 


113c— 8c+43 

6a:+18     ,^      11  — 3:r     ^        ,^     13—:?;     21— 2a; 
29.  Given -^-4-1 ^^=5.-48-^^ ^^ 

to  find  tlie  value  of  :r.  Ans.  2:=10. 

^^    ^.       4x4-3     7a;— 29     8a;+19      ,    ,   , 

oU.  Given  — f-- =-?;= — tt^ —  to  find  the  value  of  x. 

9         Dx—12         18 

Atis.     x=Q. 


SECTION    VIII. 

PROBLEMS. 

1.  A  gentleman  stated  that  his  age  was  twice  that  of  his  old- 
est son,  and  that  the  sum  of  their  ages  was  72  years.  Kequired 
the  age  of  each. 


PROBLEMS.  79 

Let  X  z=  the  age  of  the  son. 

Then  2x  =  the  age  of  the  gentleman. 

Therefore,  a;-|-2:c=72,  the  age  of  both. 

Or,  8a;=72. 

Dividing,  x=24,  the  age  of  the  son. 

2:c=48,  the  age  of  the  gentleman. 
Proof,    •  24+48=72. 

2.  What  number  is  that,  to  which  if  ^  of  it  be  added,  the 
sum  will  be  99  ? 

Let  X  =  the  number. 

4:X 

Then,  — -|-2-=99. 

Clearing  effractions,  4a;-|-7:r=693. 
Collecting  the  terms,  lla:=693. 
Dividing,  x=  63,  the  number. 

3.  A's  and  B's  estates  are  valued  at  $3240,  but  B's  is  only  | 
the  value  of  A's.     What  is  the  property  of  each  ? 

Let  X  =  A's  estate. 

7x 
Then,  -—  =  B's  estate. 

o 

Therefore,  a:+5=3240. 

o 

Clearing  effractions,      8a;+7:r=25920. 

Or,  15:^=25920. 

Dividing,  a:=:1728,    A.'s  estate. 

7X1728 

=1512,    B.'s  estate. 

o 

4.  If  §  of  a  certain  number  be  added  to  ^  of  it,  the  sum  will 
be  98.     Required  the  number. 

Let  X  =  the  number. 

Then,  T+i=^^- 

Clearing  of  fractions,  4x-{-Sx=bSS. 

Or,  7:i;=588. 

Dividing,  x=  84.     A?is. 


80  ALGEBRA. 

5.  A  certain  gentleman  divided  his  property,  consisting  of 
$5300,  among  his  four  sons.  A,  B,  C,  and  D.  He  gave  $350 
more  to  B  than  A ;  he  gave  C  $400  more  than  B ;  but  he  gave 
D  twice  as  much  as  he  gave  A  and  B.  How  much  did  each 
son  receive  ? 

Let    X  =  A's  share. 

Then  rc+350  =  B's  share. 

And   a;+ 350+400  =  C's  share. 

And   2(2z+350)=4:r+700=  D's  share. 

Therefore,  a:+a;+350-|- a: -{-850  +  400+4a;-|- 700=5300 

Reducing,  7a:+1800=5300. 

Transposing,         7a;=5300— 1800. 

Reducing,  72:= 35  00. 

Dividing,  x==  500,  A's  share. 

500+350=  850,  B's  share. 
850+400=1250,  C's  share. 
2(500+850)=2700,  D's  share. 

Verification,  500+850+1250+2700=5300. 

6.  Divide  $70  among  James,  John,  and  Charles;  give  John 
twice  as  much  as  James,  and  give  Charles  twice  as  much  as 
John. 

Let      X  =  the  sum  given  to  James. 
Then  2x  =  the  sum  given  to  John. 
And   4a;  =  the  sum  given  to  Charles. 
Then,  by  the  conditions  of  the  question, 

x-{-2xi-4:X=70. 
Or,  7:r=70. 

Dividing,  x=10,  the  sum  given  James. 

2:?;=20,  the  sum  given  John. 

42'=40,  the  sum  given  Charles. 
Verification,  10  +  20+40=70. 

7.  Two  men  found  a  purse  containing  $144,  and  it  was  agreed 
that  B  should  have  $30  more  than  A.  How  many  dollars  did 
each  receive  ? 


PROBLEMS.  81 

Let  X  =  the  sum  A  received. 
Then  x-^dO  =  the  sum  B  received. 
Therefore,  x-{-x-{~^0=lU. 
Or,  2a:+80=144. 

Transposing,  2:c=144— 30. 

Or,  2x=lU. 

Dividing,  x=  57,  the  sum  A  received. 

a:+30=  87,  the  sum  B  received. 
Verification,  57+87=144. 

8.  My  horse  and  chaise  are  worth  $336,  but  the  chaise  is 
worth  five  times  as  much  as  the  horse.  What  is  the  value  of 
each? 

Let  X  =  the  value  of  the  horse. 
Then  6x  =  the  value  of  the  chaise. 
And,  a:-f5a:=336. 

Or,  6z=336. 

Dividing,  x=  56  =  value  of  the  horse. 

5:r=280  =  value  of  the  chaise. 
Proof,  56-1-280=336. 

9.  What  number  is  that  whose  third  part  exceeds  its  fifth 
by  12? 

Let  X  =  the  number  required. 

X 

Then  its  third  part  will  be  -. 

o 

X 

And  its  fifth  part,  — . 

0 

Therefore,  ^_f=12. 

Multiplying  all  the  terms  by  15,  we  have, 

5z— 32-=180. 
Or,  22:=180. 

Dividing,  x=z  90,  the  number  required. 

10.  John  Smith's  oldest  daughter  is  15  years  old,  and  his 
youngest  daughter  is  11 ;  he  has  $1728,  which  he  wishes  to 
give  them.  How  shall  he  divide  this  sum,  that  each  may  de- 
posit her  share  in  a  bank  which  pays  6  per  cent,  simple  interest. 


82  A  L  a  E  B  K  A  . 

SO  that  both  shall  have  an  equal  sum  when  thej  are  21  years 
old? 

Let  X  =  the  sum  the  youngest  receives. 

And,  1728 — x  =  the  sum  the  oldest  receives. 

Then,  a:+a;X.06XlO=1728— a:+(1728-a:X.06x6). 

Or,  a:+.63:=1728— a:-f622.08-.36:i:. 

Transposing,  2.96:r=2350.08. 

Dividing,  a:=$793f^,  the  youngest  receives. 

$1728— $793ff =$934^-,  the  oldest  receives. 
Let  the  pupil  prove  this  question. 

11.  A  man  being  asked  the  value  of  his  horse  and  saddle, 
replied  that  his  horse  was  worth  $114  more  than  his  saddle,  and 
that  §  the  value  of  the  horse  was  7  times  the  value  of  his 
saddle.     What  was  the  value  of  each  ? 

Let  X  =  the  value  of  the  saddle. 
And  :?:-}- 114  =  the  value  of  the  horse. 
Then,       f  (:c+114)=7a:. 
Or,  2:c+228=21a; 

Transposing,        19z=228. 
Dividing,  a:=$12,  value  of  the  saddle. 

$12-{-$114=:$126,  value  of  the  horse. 

12.  A  can  reap  a  field  in  7  days,  B  can  reap  it  in  5  days. 
In  what  time  can  they  both  reap  it  together  ? 

Let  X  =  the  days  they  would  reap  it  together. 

A  would  reap  j-  of  it  in  a  day,  and  B  would  reap  -i  of  it  in  a 

1     1     12 

day ;  therefore  in  one  day  both  together  would  reap  s:+^=Qr 

of  it. 

But,  by  the  conditions,  the  field  was  to  be  reaped  in  x  days. 

Therefore,  —  :  1  :  :  1  day  :  x  days. 

oO 

12a: 
Multiplying  extremes,       -^  =  1. 

Multiplying  by  35,  12:c=35. 

Dividing,  x=  2j-^  days.     Ans. 

13.  I  have  two  carriages ;  the  value  of  one  is  five  times  that 
of  the  other,  and  the  value  of  my  horse  is  equal  to  both  of  my 


PROBLEMS.  83 

carriages.     The  worth  of  them  all  is  $300.     What  is  the  value 
of  each  ? 

Ans.  First  carriage  $25,  second  carriage  $125,  horse  $150. 

14.  A  gentleman  being  asked  his  age,  replied  that  his  was 
twice  that  of  his  wife,  and  that  his  wife  was  three  times  as  old 
as  his  daughter,  and  that  the  sum  of  their  ages  was  120  years. 
Required  the  age  of  each. 

{  Gentleman's  age,      72  years. 

Ans.  }  His  wife's  age,  36  years. 

(  His  daughter's  age,  12  years. 

15.  A  man  met  4  beggars,  to  whom  he  gave  77  cents.  To 
the  first  he  gave  twice  as  many  as  to  the  second ;  to  the  third, 
as  many  as  he  gave  to  the  first  and  second ;  and  to  the  fourth, 
as  many  as  he  gave  to  the  first  and  third.  What  sum  did  he 
give  each  ? 

Am.     First  14  cents,  second  7,  third  21,  fourth  35. 

16.  A  drover  has  a  lot  of  oxen  and  cows,  for  which  he  gave 
$1428.  For  the  oxen  he  gave  $55  each,  and  for  the  cows  $32 
each,  and  he  had  twice  as  many  cows  as  oxen.  Required  the 
number  of  each.  A7is.  12  oxen,  24  cows. 

17.  A  gentleman,  at  his  decease,  left  an  estate  of  $1872  for 
his  wife,  three  sons,  and  two  daughters.  His  wife  was  to  receive 
three  times  as  much  as  either  of  her  daughters,  and  his  sons  to 
receive  each  one  half  as  much  as  one  of  the  daughters.  Re- 
quired the  sum  each  received. 

Atis.  Wife  $864,  daughters  $288  each,  sons  $144  each. 

18.  A  boy  bought  apples,  oranges,  and  pears ;  he  gave  two 
cents  a-piece  for  the  apples,  three  cents  for  the  oranges,  and 
four  cents  for  the  pears.  He  had  twice  as  many  oranges  as 
apples,  and  three  times  as  many  pears  as  oranges.  The  sum  he 
expended  was  $2.24.     How  many  did  he  buy  of  each  kind  ? 

Ans.  7  apples,  14  oranges,  42  pears. 

19  Let  85  be  divided  into  two  such  parts  that  one  of  them 
shall  be  four  times  as  large  as  the  other.         A71S.  17  and  68. 


84  ALGEBRA. 

20.  Divide  $100  among  A,  B,  and  C,  so  that  A  may  have 
$20  more  than  B,  and  B  $10  more  than  C. 

Am.     A  $50,  B  $30,  and  C  $20. 

21.  A  prize  of  $1000  is  to  be  divided  between  A  and  B,  so 
that  their  shares  may  be  in  the  proportion  of  7  to  8 ;  required 
the  share  of  each.         Ans.  A's  share  $4G6|,  and  B's  $533^. 

22.  What  number  is  that  whose  3d  part  exceeds  its  5th  part 
by  6f  ?  A71S.     48. 

23.  A  laborer  agreed  to  serve  for  36  days  on  these  condi- 
tions; that  for  every  day  he  worked  he  was  to  receive  $1.25, 
but  for  every  day  he  was  absent  he  was  to  forfeit  $0.50.  At 
the  end  of  the  time  he  received  $17.  It  is  required  to  find 
how  many  days  he  labored,  and  how  many  days  he  was  absent. 

A71S.     He  labored  20  days,  and  was  absent  16  days. 

24.  Out  of  a  cask  of  wine,  which  had  leaked  away  ^,  13 
gallons  were  drawn,  and  then  being  gauged  it  was  found  to  be 
half  full.     How  many  gallons  did  the  cask  contain  ? 

Ans.     78  gallons. 

25.  Divide  30  into  two  such  parts  that  f  of  the  one  shall 
exceed  f  of  the  other  by  6f.  Atis.     18  and  12. 

26.  Wliat  two  numbers  are  those  whose  difference  is  3,  and 
the  difference  of  whose  squares  is  51  ?  Atis.     10  and  7. 

27.  Three  men,  A,  B,  and  C,  trade  in  company  ;  A  put  in  a 
certain  sum,  B  put  in  twice  as  much  as  A,  and  C  put  in  three 
times  as  much  as  both,  and  they  gain  $864.  What  is  each 
man's  share  of  the  gain  ? 

Ans.     A's  $72,  B's  $144,  C's  $648. 

28.  James  and  William  have  between  them  44  apples,  and 
James  says  to  William,  if  you  will  give  me  12  of  your  apples, 
your  number  will  then  be  only  f  of  mine.  William  replied,  if 
you  will  give  me  12  of  yours,  your  number  will  then  be  only  § 
of  mine.     Required  the  number  of  each. 

Ans.     James  had  24  apples,  and  William  20. 


—/^ 


PROBLEMS.  85 

29.  Let  112  be  divided  into  two  such  numbers  that  the 
gi'eater  shall  be  to  the  less  as  9  to  7.  Ans.     G3  and  49. 

30.  Let  19  be  divided  into  two  such  parts  that  three  times 
the  greater  shall  be  equal  to  four  times  the  less.  Required 
those  numbers.  Ans.     lOf  and  8|. 

31.  There  are  two  numbers  whose  sum  is  24,  and  if  7  be 
added  to  the  larger,  and  4  to  the  less,  their  ratio  will  be  as  4  to 
3.     Required  those  numbers.  Ans.     13  and  11. 

32.  The  difference  of  two  numbers  is  4,  and  7  times  the 
larger  number  is  equal  to  11  times  the  less.  Required  those 
numbers.  Ans.     11  and  7. 

33.  A  merchant  has  two  kinds  of  grain,  one  at  $2.50  per 
bushel,  and  the  other  at  S2  per  bushel.  He  wishes  to  make 
a  mixture  of  80  bushels,  that  shall  be  worth  $2.10  per  bushel. 
How  mjanj  bushels  of  each  sort  must  he  use  ? 

'"'  "'/^  ^"  ^       '  Ans.     16  bushels  at  $2.50,  and  64  at  $2. 

34.  A  man  having  lost  ^  of  his  money,  found  he  had  $96 
left.     Required  the  sum  he  had  at  first.  An^.     $128. 

35.  J.  Jones  found  a  certain  sum  of  money,  which  was  equal 
to  ^  of  what  he  possessed ;  but  having  spent  $40,  the  remainder 
was  f  of  the  sum  he  found.  Required  the  sum  he  at  first 
possessed.  -f^-d^-Oz.  %2  Ans.     $36f_. 

36.  In  my  school  |  of  my  pupils  study  grammar,  §  of  the 
remainder  read,  10  spell,  and  the  remainder,  which  is  ^  of  the 
number  that  read,  study  navigation.  Required  the  number  of 
pupils  in  the  school.  A')is.     70  pupils. 

37.  A  gentleman  lent  a  certain  sum  of  money  for  3  years  at' 
5  per  cent,  compound  interest ;  that  is,  at  the  end  of  each  year 
he  added  ^'^  to  the  sum  due.     At  the  close  of  the  third  year  he 
lost  $15.25,  but  then  there  remained  due  to  him  $2300.     Re- 
quired the  sum  lent.  A^is.    82000. 

38.  A  spendthrift  spent  -i-  of  the  fortune  left  him  by  his 
father,  and  he  then  earned  $124.  Soon  after  he  lost  in  specu- 
lation §  of  his   property,  after  which   he   gained  $274.     His 

8 


86  ^  ALGEBEA. 

property  was   now  valued   at  ^,  wanting  $86,  of  his  original 
estate.     What  was  the  sum  left  him  by  his  father  ? 

Am.     $1720. 

39.  A  asked  B  how  much  money  he  had.  He  replied,  if  I 
had  5  times  the  sum  I  now  possess  I  could  lend  you  $60,  and 
then  -1-  of  the  remainder  would  be  equal  to  ^  the  dollars  I  now 
have.     Required  the  sum  which  B  had.      "■"  ^  ^.    A^is.     $24. 

40.  A  gentleman  left  an  estate  of  $1862  for  his  three  sons. 
He  gave  his  youngest  $133  less  than  his  second  son,  and  to  his 
oldest  son  he  gave  as  much  as  to  the  other  two.  How  much 
did  each  receive  ? 

Ans.     Youngest  son  $399,  second  $532,  oldest  $931. 

41.  A,  B  and  C,  found  a  purse  of  money,  and  it  was  mutu- 
ally agreed  that  A  should  receive  $15  less  than  one-half,  and 
that  B  should  have  $13  more  than  one  quarter,  and  that  C 
should  have  the  remainder,  which  was  $27.  How  many  dollars 
did  the  purse  contain  ?  .      Ans.     $100. 

42.  Lent  my  good  friend  S.  Jenkins  a  certain  sum  of  money, 
at  6  per  cent.,  which  he  kept  until  the  interest  was  ^  of  the 
principal.  The  sum  then  due  was  $500.  Required  the  sum 
lent.  A71S.     $350. 

43.  A  certain  man  added  to  his  estate  \  its  value,  and  then 
lost  $760.  But  he  afterwards  gained  $600.  His  property  then 
amounted  to  $2000.     What  was  the  value  of  his  estate  at  first  ? 

Ans.     $1728. 

44.  James  said  to  John,  I  have  40  shillings  more  than  you. 
Yes,  replied  the  other,  and  ^  of  yours  is  equal  to  \  of  mine. 
Required  the  number  of  shillings  that  each  had. 

Arts.     James  72  shillings,  and  John  32. 

45.  A  merchant  bought  a  number  of  barrels  of  flour,  and 
having  sold  half  the  number  and  4  barrels  more  to  A,  and  |-  of 
the  remainder  wanting  4  barrels  to  B,  he  had  20  barrels  re- 
maining.    Required  the  number  the  merchant  bought. 

Ans.     1 36  barrels. 


//> 


PROBLEMS.  87 

46.  What  number  is  that  from  which,  if  7  be  subtracted,  ^ 
of  the  remainder  will  be  5  ?  Ans.     37. 

47.  It  is  required  to  divide  44  into  two  such  numbers  that  | 
of  one  of  them  shall  be  6  more  than  f  of  the  other. 

.^  -  /:  -n    ^  /  k  I'  -.  %  Ans.     24  and  20. 


48.  It  is  required  to  divide  the  number  43  into  two  such 
parts  that  one  of  them  shall  be  3  times  as  much  above  20  as  the 
other  wants  of  17.     Required  the  numbers. 

'  /'  Ans.     29  and  14. 

49.  John  Jones  can  reap  a  certain  field  in  10  days,  but,  with 
the  help  of  his  oldest  son,  he  can  do  it  in  8  days.  How  long 
would  it  require  his  son  to  perform  the  labor  himself  ? 

Ans.     40  days. 

50.  A  engaged  to  reap  a  field  for  90  shillings,  and  he  could 
perform  the  labor  in  9  days  ;  but  he  took  in  B  as  a  partner,  and 
they  supposed  it  would  require  5  days  for  both  to  perform  the 
labor,  but  they  finished  it  in  4  days.  How  much,  in  justice, 
must  A  pay  to  B  ?  Ans.     50  shillings. 

'tJl.  I  have  two  horses,  and  a  saddle  worth  $30.  Now,  the 
saddle  and  first  horse  are  worth  f  the  second  horse,  but  the 
saddle  and  second  horse  are  worth  three  times  the  first  horse. 
Required  the  value  of  each. 

Ans.     First  horse  $60,  second  horse  $150. 

52.  A  gentleman  let  f  of  his  money  at  5  per  cent.,  and  the 
fv,  remainder  at  6  per  cent.,  and  his  interest  amounted  to  $180. 

What  were  the  sums  lent  ? 

Am.     $1200  at  5  per  cent.,  $2000  at  6  per  cent. 

53.  A  can  do  a  piece  of  work  in  12  days,  B  can  do  the  same 
work  in  10  days,  and  C  can  perform  it  in  8  days.  How  long 
would  it  require  A  and  B  to  do  it ;  how  long  A  and  C ;  how 
long  B  and  C ;  and  how  long  A,  B  and  C,  to  perform  the  labor  ? 

An^.  A  and  B  b^^  days,  A  and  C  4|  days,  B  and  C  4|- 
days,  A,  B  and  C,  3/^  days. 


OO  ALGEBRA. 

54.  Lent  $TSO,  at  6  p^r  cent ,  for  5  rears.  Wlmt  principal 
will  aniomit  to  the  sum  in  4  years,  at  10  per  cent  ? 

Ans,  $724,284. 

55.  Lent  mj  neighbor  Jenkins  8270  for  4  Tears,  at  6  per 
cent, ;  some  time  afterwards.  I  borrowed  of  him  $500,  at  S  per 
cent.     How  long  shall  I  keep  it,  to  balance  the  faTor  ? 

-z:^^  K  S-x  -^  ^^  7«?  X  ^K '       Ans.  1|^  years, 

b'o.  A  fox  is  pnrsned  by  a  greyhoimd,  and  is  60  of  her  own 
leaps  before  him.  The  fox  makes  9  leaps  while  the  greyhoiind 
makes  bnt  6 :  but  the  latter  in  3  leaps  goes  as  far  as  the  former 
in  7.  How  many  leaps  does  the  greyhound  make  before  he 
catches  the  fox  ? 

Ans.  The  greyhonnd  makes  72  leaps,  and  the  fox  lOS. 

57.  A  gentleman  gave  in  charity  $-r6  :  a  part  thereof  in  eqnal 
portions  to  five  poor  men,  and  the  rest  in  equal  portions  to  7 
poor  women,  Xow.  a  man  and  a  woman  had  between  them  $8. 
What  was  given  to  the  men.  and  what  to  the  women  ? 

Ans.  The  men  receired  $25,  and  the  women  $2L 

5S.  A  man  has  two  farms,  and  his  stock  is  worth  $18o.  !Now, 
the  stock  and  his  first  farm  is  worth  once  and  two-sevenths  the 
value  of  the  second  farm,  but  the  stock  and  the  second  farm  is 
worth  once  and  five-eiirhths  the  value  of  the  first  faim.  "What  is 
the  value  of  each  &rm  ? 

Atis.  First  farm,  $384;  second  farm,  $441. 

59.  A  certain  clock  has  an  hour  hand,  a  minute  hand,  and  a 
second  hand,  all  turning  on  the  same  centre.  At  12  o'clock 
all  the  hands  are  together,  and  point  at  12,  How  long  will  it 
be  before  the  second  hand  will  be  between  the  other  two  hands, 
and  at  equal  distances  from  each  ?  Also,  before  the  minute 
hand  will  be  equally  distant  between  the  other  two  hands  ? 
Also,  before  the  hour  hand  will  be  equally  distant  between  the 
other  two  hands  ? 

Jjis,  60^^^j  seconds,  i31f|4  seconds,  594-f  seconds. 

GO.  TThat  ntimber  is  that,  the  treble  of  which,  increased  by 
12,  t'^hnll  as  much  exceed  54  as  that  treble  is  less  than  144  ? 

Ans.  31. 


EQUATIONS 


89 


SECTION    IX. 

EQUATIONS    OF   THE   FIRST   DEGREE,    CONTAINING   TWO 
UNKNOWN   QUANTITIES. 

Art.  149.  When  the  problem  contains  two  unknown  quantities, 
there  must  be  two  independent  equations  involving  them ;  and 
from  them  an  equation  may  be  deduced,  which  shall  contain 
only  one  of  the  unknown  quantities. 

The  process  by  which  one  of  the  unknown  quantities  is  thus 
removed  is  called  elimination;  and  this  may  be  performed 
in  three  ways. 

First,  by  Addition  and  Subtraction. 

Second,  by  Comparison. 

Third,  by  Substitution. 

150.  Elimination  by  addition  and  subtraction. 


EXAMPLES. 


1.  Given    j  p     ,  ;:  PY  (  to  find  the  value  of  a;  and  y. 


1.  By  first  condition, 

2.  By  second      " 

3.  Multiplying  1st  by  2, 

4.  Multiplying  2d  by  1, 

5.  Subtracting  3d  from  4th, 

6.  Dividing  5th  by  9, 

7.  Multiplying  1st  by  5, 

8.  Multiplying  2d  by  2, 

9.  Adding  7th  and  8th, 
10.  Dividing  9th  by  27, 

VEKIFICATION. 


Sx—2y=  11. 
Qx-\-^y=  67. 
6a;— 47/=  22. 
Qx-{-5y=  67. 
9y=z  45. 

y=    5- 

lDz—10y=  55. 

12a:+10z/=134. 

272:=189. 

x=     7. 


3x7—2x5=21-10=11. 

6x7+5x5=42-1-25=67. 
8^ 


90  ALGEBRA. 


(  5:r-[-42/=23  ) 
2.  Given    |  Qx—Sv=12  )  *^  ^^^  *^^  value  ofx  and  ?/. 


1. 

By  the  first  condition, 

bx-\-4:y=  23. 

2. 

By  the  second, 

Qx—By=  12. 

3. 

Multiplying  1st  by  6, 

302;+24?/=:138. 

4. 

Multiplying  2d  by  5, 

30^—152/=  60. 

5. 

Subtracting  4th  from  3d, 

39?/=  78. 

6. 

Dividing  5th  by  39, 

y=    2. 

7. 

Multiplying  1st  by  3, 

15x-\-12y=  69. 

8. 

Multiplying  2d  by  4, 

24:x—12y=  48. 

9. 

Adding  7th  and  8th, 

39a:=117. 

10. 

Dividing  9th  by  39, 

VERIFICATION. 

x=     3. 

5x3+4x2=15+8=23. 

6x3—3x2=18-6=12. 
3.  A  says  to  B,  if  \  of  my  age  were  added  to  |  of  yours,  the 
sum  would  be  19^  years.     But,  says  B,  if  f  of  mine  were  sub- 
tracted from  I  of  yours,  the  remainder  would  be  18^  years. 
Required  the  sum  of  their  ages. 

X     2?y 
1.  By  first  condition,  -+  -^=       194-. 

5      o 


2.  By  the  second,  -^ /=       18^. 


7x_2y 

3.  Clearing  the  1st  of  fractions,        Sx-{-10y=  290. 

4.  Clearing  the  2d,                           35a:— 162/=  730. 

5.  Multiplying  3d  by  35,  105a;+350?/=10150. 

6.  Multiplying  4th  by  3,              105a;—  48?/=  2190. 

7.  Subtracting  6th  from  5th,                   398?/=  7960. 

8.  Dividing  7th  by  398,                                 y=  20. 

9.  Substituting  20  for  y  in  the  3d,    3a;+200=  290. 

10.  Transposing  and  uniting,  Sx==       90. 

11.  Dividing  10th  by  3,  x=z      30. 

VERIFICATION. 

30  .  2X20 


5^     3 


=6+13^=19^ 


TT* 


I><L0_?><20^2Gi-8=18i. 


EQUATIONS.  91 

From  the  operation  of  the  preceding  examples,  we  deduce  the 
following 

Rule.  Multiply  or  divide  the  given  equations  by  such  num- 
bers or  quantities  as  will  make  ike  term  that  contains  one  of  the 
unkrwwn  quantities  the  same  in  each  of  them;  then  add  or 
subtract  the  tivo  equations  thus  obtained^  and  there  will  arise  a 
new  equation  with  only  one  unknown  quantity  in  it,  which  may 
be  resolved  by  Art.  147. 

151 1  Elimination  by  comparison. 

4.  Given    5  ^   """^^     -  .  [  to  find  the  values  of  a:  and  y. 
(  bz — z?/=14  ) 

1.  By  the  first  condition,  1x-\-^y=Vl. 

2.  By  the  second,  5a;— 2?/=14. 

3.  Transposition  of  the  1st,  2x=.VJ—^y. 

4.  Dividing  the  3d  by  2,  x= — ^^. 

5.  Transposition  of  the  2d,  bx=:l^-\-2y. 

14+2?/ 

6.  Dividing  the  5th  by  5,  x= — i— ^. 

As  things  which  are  equal  to  the  same  are  equal  to   each 
other,  we  therefore  infer  that  — ^— ^,    in  the  4th,  is  equal  to 

— ^^—^  in  the  6th ;  because  they  are  both  equal  to  x. 

0 


7. 

Therefore, 

17 

-2>y     14+22/ 
2               5      • 

8. 

Clearing  of  fractions, 

85- 

-152/=28+42/. 

9. 
10. 

Transposing  8th, 
Dividing  9th  by  19, 

192/=57. 
2/=3. 

11. 

Substituting  3  for  the  value 

ofy 

in 

the  first  equation,  we 

have, 

22:=17-9. 

12. 

By  reduction, 

2a;=8. 

13. 

Dividing  12th  by  2, 

a;=4. 

92  ALGEBRA 

VERIFICATION. 

2x4+3x3=  8-f9=17. 
5x4-2x3=20-6=14. 
Hence  the  following 

EuLE.  Observe  which  of  the  unknown  quantities  is  least  m- 
volved^  and  find  its  value  in  each  of  the  equations,  as  i?L 
Art.  148. 

Let  the  two  values  thus  found  be  made  equal  to  each  other,  and 
there  will  arise  a  new  equation,  with  only  one  unkrwwn  quantity 
in  it,  whose  value  may  be  found  as  in  Art.  147. 

152i  Elimination  by  substitution. 

5.  Two  boys  playing  marbles,  the  older  said  to  the  younger, 
if  you  had  three  times  as  many  marbles  as  you  now  possess, 
the  sum  of  yours  and  mine  would  be  19.  But  the  younger 
replied,  if  twice  the  number  of  mine  were  subtracted  from  four 
times  as  many  as  you  have,  the  number  would  be  20.  Required 
the  number  of  marbles  that  each  possessed. 

Let  X  represent  the  marbles  of  the  elder ; 
And  y  the  number  of  the  younger. 

1.  Then,  by  the  condition  of  the  question,      x-\-Zy=\^. 

2.  And  4a:— 2y=20. 

3.  Transposing  the  1st,  :r=19 — 'dy 

4.  Putting  the  3d  into  the  2d,  4(19— 3?/)— 22/=20. 

5.  Then,  76— 12?/— 22/=20. 

6.  Transposing  and  reducing,  2/=4. 

7.  Putting  the  value  ofy  into  the  1st,  :c-|-12=19. 

8.  Transposing  and  reducing,  a:=19 — 12=7. 

Alls.    The  elder  had  7  marbles,  and  the  younger  4. 

VERIFICATION. 

7+3x4=  7+12=19. 
4X7-2X4=28-  8=20. 

By  the  above  method  of  operation,  we  deduce  the  following 

BuLE.  Find  the  value  of  either  of  the  unknown  quantities  iii 
that  equation  in  lohich  it  is  least  involved  ;  then  substitute  this 
value  in  the  place  of  its  equal  in  the  other  equation,  and  there 


EQUATIONS.  93 

will  arise  a  new  equation,  with  only  one  unknown  quantity  in 
it ;  the  value  ofivhich  may  he  found  by  Art.  147. 

EXAMPLES. 

Ans.  2::=4;  y=^. 

7.  Given    <  ^  ^  J      Required  x  and  y. 

Ans.  a:=5 ;  y-=-1. 

8.  Given    \  }      Required  a;  and  y. 

Ans.     x=.l ;  y=^^. 

9.  Given    <  o  _or  \      I^equired  x  and  ?/. 

Atis.     x=S  :  ?/=2. 

_„     „.  (  12x-}-Sy=llQ)      T,       .     , 

10.  Given    jo  o  \      Required  z  and  y. 

Ans.    x=6  ;  y=7. 

^^    ^.  {  llx-\-3y=  124) 

11.  Given    i    ^       n  rr>{    to  find  X  and  ?/. 

(    2x—Qy=z—oQ  )  ^ 

Ans.     a:=8;  y=Vl. 

-lo    n-  (9a:+42/=58)        ^   , 

12.  Given    s  «    '  ^^  f   to  find  x  and  ?/. 

(  32;-|-2?/=26  )  ^ 

^?w.     a;=2;  ?/=10. 

13.  Given    ]  „  ^       o^^  c   to  find  the  value  of  a;  and  y. 

(  ^x—1y=  80  )  -^ 

Ans.     a:=12;  y=&. 

(  lx—2y= 6  ) 

14.  Given    <  ^    ,  _         r.  ^  c   to  find  the  value  of  x  and  y. 

(  2x-{-2y=i  24  )  ^ 

Ans.     x=2;  2/=10. 

15.  Given    \  ^       „^  o„  [   to  find  the  value  ofx  and  y. 

(  b^;— 222/=  — 30  )  ^ 

^?z^.     a:=10;  ?/=5. 

(    2x4-  3?/=     47  ) 

16.  Given    \  '       -^  ^.^  C   to  fiii<i  the  value  of  x  and  y. 

(  10:c— 12?/=— 62  )  '^ 

Ans.    x=7 ;  y=ll. 


94 


ALGEBRA. 


17.  Given 


18.  Given 


19.  Given 


20.  Given 


21.  Given 


2     3 
4^6 


(Sx 

a:+|=37 
'  5 


(4x_2y_       1 
7       3"" 

[  a:+7^=175j 


8      9" 


to  find  the  value  of  x  and  y. 

Am.     a;=12;  2/=18. 

•  to  find  the  value  of  x  and  y. 

Ans.     a;=35;  y='^^. 
•  to  find  the  value  of  a:  and  y. 
Ans.     a:=28;  2/=21. 


19 


■  to  find  the  value  of  x  and  y» 


3:c+32/=126^ 

Ans.    a:=24;  2/=18. 

'U:c+^y      38] 

t>  V  to  find  the  value  of  x  and  y. 


[  x-{-12y=UQ\ 


Ans.     x=2;  y=12. 


22.  Given 


23.  Given 


r^_7^_ 

7     10"~ 


X 


;+32/=134 

4  '    ^  i 


►  to  find  the  value  of  x  and  y. 


Ans.     a;=56;  y=4:0. 

!,  ,  f  to  find  the  value  of  x  and  ?/. 

mx-\-ny=d  ) 


.  3^ — 7ZC  «^ — 77ZC 

Ans.     x=- ;  y= 1— . 

om — an  an — om 


24c.  Given 


a     h 


:in 


\-Y-=n 
\c     a 


■  to  find  X  and  y. 


.                ahcm-\-acdn  bcdn — abdm 

Ans.    x= :rT-7-- — ;  y= 


ad-\-bc 


ad-\-bc 


EQUATIONS 


95 


25.  Given 


[|-12=|+8 

5    ^3  4     ^ 


to  find  the  value  of 
X  and  ?/. 


Am.     a:=60;  2/=40. 


SECTION   X. 

ELIMINATION  WHERE  THERE  ARE  THREE  OR  MORE  UN- 
KNOWN QUANTITIES  INVOLVED  IN  AN  EQUAL  NUMBER 
OF   EQUATIONS. 

Rule.  Find  the  values  of  one  of  the  unhioivn  quantities  in 
each  of  the  three  given  equations,  as  if  all  the  others  ivere 
knmjvn  ;  then  put  the  first  of  these  values  equal  to  the  second, 
and  either  the  first  or  second  equal  to  the  third,  and  there  ivill 
arise  two  neio  equations  with  only  two  unhioivn  quantities  in 
them,  the  values  of  which  may  he  found  as  in  Art.  147,  and 
thence  the  value  of  the  third. 

Or,  the  unknown  quantities  may  be  obtained  by  multiplying 
each  of  the  three  equations  by  such  quantities  as  ivill  make  one 
of  their  terms  the  same  iji  all  of  them ;  then,  having  sub- 
tracted any  two  of  these  resulting  equations  from  the  third,  or 
added  them  together,  as  the  case  may  require,  there  will  remain 
only  two  equations,  which  may  be  resolved  by  the  former  rules. 

Or,  toe  may  find  the  value  of  one  of  the  unknown  quantities  in 
that  equation  in  which  it  is  least  involved,  and  then  substitute 
this  value  for  that  unknown  quantity  in  all  the  other  equxitions, 
and,  proceeding  in  the  same  way  with  these  equations,  we  obtain 
the  other  unknown  quantities. 


1.  Griven 


EXAMPLES. 

a:-{-37/-[-  z=47 


-+  ^4-  -=10 

2^  3^  4 


>  to  find  the  value  of  x,  y  and  z. 


96  ALGEBRA. 

4.  From  the  1st  equation,  x=41 —  y — 2z. 

5.  From  the  2d,  x=4:7—Sy—  z. 

6.  From  the  3d,  ^,-=20—?^-  -. 

3       2 

7.  Equal  values  of  x  in  4th  and 

5th,  41— 2/— 2z=47— 37/— z. 

8.  Value  ofy  in  7th,  2/=^i^- 

9.  Equal   values   of  x   in   4th   and 

6th,  41-2/-2z=20-?^-^. 

^  3      2 

10.  Value  of  2/  in  9th,  2/=63-^. 

11.  Equal  values  of  2/  in  8th  and  10th,  ^±5=63——. 

12.  Keducing,  z=12. 

13.  Substituting  for  z  its  value  in  8th,  ?/= — - — =9. 

14.  Substituting  for  y  and  z  their  values 

in  4th,  2:=41— 9— 24=8. 

1  /    5a:+4?/-2z=28  \       ^   ^    , 

on-        o)in       o     \  A       on  (to  nnd  the  value  of  x,  y, 
2.  Given  2  <  10a: — 6?/+4z=30  >  ,  -^ 

Q  /    o     .  a  \      and  2r. 

3  \    Zx-\-  y —  z=  9  ; 

Subtracting  the  2d  from  twice  the  1st,  we  have, 

4.  142/— 8z=26. 

Subtracting  the  2d  from  5  times  the  3d, 

5.  Il2/-9z=15. 

Subtracting  14  times  the  5th  from  11  times  the  4th, 

6.  382=76. 

7.  2=2. 

Substituting  for  z  its  value  in  the  5th, 

8.  Il2/-18=15. 

9.  ?/=3. 


EQUATIONS. 


97 


Substituting  for  y  and  z  their  values  in  the  3d, 

10.  2.T-f  3— 2=9. 

11.  a:=4. 

{  Zx—  y-2z=  0  ^ 

3.  Given   <  6z-}-2?/-|-3z=45  >  to  find  x,  y,  and  z. 

I  4:X-\-Sy—  ^=-31  ) 

A71S.  2:=4;  y=Q;  z=o. 

/    8a;-%-7z=-36  ^ 

4.  Given  <  12:c —  ?/— 3z=     36  >  to  find  x,  y,  and  z. 

(    Qx—2y—  z=     10  ) 

A71S.  x=4:;  y=G',  z=2. 

r  7x-\-4:y—  z=     7S  \ 

5.  Given  <  4x — by—Szz= — 21  >  to  find  x,  y,  and  z. 

(    a:— 3z/-42=— 37  ) 

Am.  x=S;  y=:7 ;  z=Q. 

/  x^y=SO  \ 

6.  Given  <  x-\-z=2b  >  to  find  x,  y,  and  z. 

(  y-i-z=W  ) 

Am.  a:=20;  y=\^\  zz=b. 

(  Sx—4:y=:24:—  z  \ 

7.  Given   <  6a: -j-  y=  2:-{-84  >  to  find  x,  y,  and  z. 

(    a;-f80=32/+4z  ) 

^7w.  a:=12;  2/=20;  z=8. 

f  ,^_^=23" 
2^3     4 

8.  Given  -j  -—=^-4 — =12  \-io  find  x,  y,  and  2r. 
3    4^  2  ' 

U^2     3 

Am.  a;=36;  2/=24;  z=12. 
C3w-f  x-{-2y—  z=22^ 


J  4^;—  ?/+3z=35 

9.  Given  i   .     ,  ^       /  .^  ^ 

4tu-\-2»x—2y         =19 

Uw         _}-42/-f-2z=46 


to  find  z^,  a:,  ?/,  and  z. 
yl?;„«.   7/=4;  2;=:5  ;  ?/=:6  ;  z=7. 


98  ALGEBRA. 


EQUATIONS   OF    THE   FIRST   DEGREE,    CONTAINING    SEYERAL   UN- 
KNOWN   QUANTITIES. 

EXAJSIPLES. 

1.  A  says  to  B  and  C,  give  me  half  of  your  money,  and  I 
shall  have  $55.  B  replies,  if  you  two  will  give  me  one  third 
of  yours,  I  shall  have  $50.  But  C  says  to  A  and  B,  if  I  had 
one  fifth  of  your  money,  I  should  have  $50.  Required  the  sum 
that  each  possessed. 

Am.  A=$20,  B=$30,  C=$40. 

2.  A  merchant  has  three  kinds  of  sugar.  He  can  sell  3  lbs. 
of  the  first  quality,  4  lbs.  of  the  second  quality,  and  2  lbs.  of 
the  third  quality,  for  60  cents ;  or,  he  can  sell  4  lbs.  of  the  first 
quality,  1  lb.  of  the  second  quality,  and  5  lbs.  of  the  third 
quality,  for  59  cents ;  or,  he  can  sell  1  lb.  of  the  first  quality, 
10  lbs.  of  the  second  quality,  and  3  lbs.  of  the  third  quality, 
for  90  cents.     Required  the  price  of  each  quality. 

A71S.  First  quality,  8  cents  per  lb. ;  second,  7  cents ;  third,  4 
cents. 

3.  A  gentleman's  two  horses,  with  their  harness,  cost  him 
$120.  The  value  of  the  worst  horse,  with  the  harness,  was 
double  that  of  the  best  horse ;  and  the  value  of  the  best  horse, 
with  the  harness,  was  triple  that  of  the  worst  horse.  What  was 
the  value  of  each  ? 

Atis.   Harness,  $50 ;  best  horse,  $40 ;    worst,  $30. 

4.  Find  three  numbers,  so  that  the  first  with  half  the  other 
two,  the  second  with  one  third  of  the  other  two,  and  the  third 
with  one  fourth  of  the  other  two,  shall  each  be  equal  to  34. 

A71S,  10,  22,  and  26. 

5.  Find  a  number  of  three  places,  of  which  the  digits  have 
equal  difi"erences  in  their  order ;  and,  if  the  number  be  divided 
by  half  the  sum  of  the  digits,  the  quotient  will  be  41 ;  and,  if 
396  be  added  to  the  number,  the  digits  will  be  inverted. 

A?is.    246. 


EQUATIONS.  99 

6.  A  farmer  has  a  large  box,  filled  with  wheat  and  rye ;  seven 
times  the  bushels  of  wheat  is  equal  to  four  times  the  bushels  of 
rye,  wanting  3  bushels ;  and  the  quantity  of  wheat  is  to  the 
quantity  of  rye  as  3  to  5.  Required  the  bushels  of  wheat  and 
the  bushels  of  rye. 

Ans.  Wheat  9  bushels,  rye  15  bushels. 

7.  A  says  to  B,  if  7  times  my  property  were  added  to  -f  of 
yours,  the  sum  would  be  $990.  B  replied,  if  7  times  my  prop- 
erty were  added  to  ^  of  yours,  the  sum  would  be  $510.  Re- 
quired the  property  of  each.  Atis.  A's,  $140 ;  B's,  $70. 

8.  If  j-  of  A's  age  were  subtracted  from  B's  age,  and  5 
years  added  to  the  remainder,  the  sum  would  be  6  years ;  and 
if  four  years  were  added  to  -i  of  B's  age,  it  would  be  equal  to 
-j^L  of  A's  age.     Required  their  ages. 

Atis.  A's,  98  years ;  B's,  15  years. 

9.  "What  fraction  is  that,  if  1  be  added  to  its  numerator,  its 
value  is  ^ ;  or,  if  1  be  added  to  its  denominator,  its  value  is  |-  ? 

Am.  /^. 

10.  A  says  to  B,  if  ^  the  difference  of  our  ages  were  sub- 
tracted from  my  age,  the  remainder  would  be  25  years.  B 
replies,  if  ^  of  the  sum  of  our  ages  were  taken  from  mine,  the 
remainder  would  be  -^  of  yours.     Required  their  ages. 

Ans.   A's,  30  years;  B's,  20  years. 

11.  There  are  two  numbers,  and  if  ^  of  their  difference  were 
taken  from  4  times  their  sum,  the  remainder  would  be  62 ;  but 
the  difference  of  their  sum  and  difference  is  equal  to  f  of  the 
larger  number.     Required  the  numbers.  Atis.  12  and  4. 

12.  Three  men  reckoning  their  money,  says  the  first,  if  $100 
were  added  to  my  money,  it  would  be  as  much  as  you  both 
possess.  Says  the  second,  if  $100  were  added  to  my  money,  I 
should  have  twice  as  much  as  you  two  have.  Says  the  third 
man,  if  $100  were  added  to  mine,  I  should  have  three  times  as 
much  as  you  both  have.     How  much  money  had  each  man  ? 

Am.  First,  $dj\,  second,  $45y\,  third,  $63^^^. 

13.  A,  B  and  C,  speaking  of  their  ages,  A  said  that  the  sum 


100 


ALGEBRA. 


of  their  ages  was  90.  B  replied,  that  if  his  age  were  taken 
from  the  sum  of  the  other  two,  the  remainder  would  be  30.  G 
said,  if  his  age  were  taken  from  the  other  two,  the  remainder 
would  be  ^  his  age.     Required  their  ages. 

Am.  A's,  20;  B's,  30;  C's,  40. 

14.  There  are  4  men,  A,  B,  C  and  D,  the  value  of  whose 
estate  is  $14,000 ;  twice  A's,  three  times  B's,  half  of  C's,  and 
one  fifth  of  D's,  is  $16,000 ;  A's,  twice  B's,  twice  C's,  and 
two  fifths  of  D's,  is  $18,000 ;  and  half  of  A's,  with  one  third 
of  B's,  one  fourth  of  C's,  and  one  fifth  of  D's,  $4000.  Re- 
quired the  property  of  each. 

Ans.  A's,  $2000;  B's,  $3000;  C's,  $4000;  D's,  $5000. 

15.  Find  four  numbers,  such  that  the  first,  together  with  half 
the  second,  may  be  357  ;  the  second,  with  4-  of  the  third,  equal 
to  476 ;  the  third,  with  ^  of  the  fourth,  equal  to  595  ;  and  the 
fourth,  with  -i-  of  the  first,  equal  to  714. 

Ans.  First  number,  190;  second,  334;  third,  426;  fourth,  676. 

16.  If  I  were  to  enlarge  my  field  by  making  it  5  rods  longer 
and  4  rods  wider,  it  would  contain  240  square  rods  more  than 
it  now  does ;  but,  if  I  were  to  make  its  length  4  rods  less,  and 
its  breadth  5  rods  less,  its  contents  would  be  210  square  rods 
less  than  its  present  surface.  What  are  its  present  length, 
breadth,  and  contents  ? 

A71S.  Length,  30  rods ;  breadth,  20  rods ;  contents,  60-0  square 
rods. 

17.  A  person  exchanged  12  bushels  of  wheat  for  8  bushels 
of  barley,  and  £2  16^. ;  ofiering,  at  the  same  time,  to  sell  a 
C3rtain  quantity  of  wheat  for  an  equal  quantity  of  barley,  and 
£3  155.  in  cash,  or  for  £10  in  cash.  Required  the  prices  of  the 
wheat  and  barley  per  bushel. 

Ans.  Wheat  at  8  shillings,  barley  at  5  shillings,  per  bushel. 

18.  A  farmer,  having  89  oxen  and  cows,  found,  after  he  had 
sold  4  oxen  and  20  cows,  he  had  7  more  oxen  than  cows.  What 
number  had  he  of  each  at  first  ?      Ans.  40  oxen  and  49  cows. 

19.  A  and  B  driving  their  turkeys  to  market,  A  sa^s  to  B, 


EQUATIONS.  101 

give  me  5  of  your  turkeys,  and  I  shall  have  as  many  as  you. 
B  replies,  but  give  me  15  of  yours,  and  then  yours  will  be  f  of 
mine.     What  number  of  turkeys  had  each  ? 

Ans.  A  45  and  B  55  turkeys. 

/  20.  It  is  required  to  find  two  such  numbers,  that  if  i  of  the 
first  be  added  to  -^  of  the  second,  the  sum  shall  be  25 ;  but,  if  ^ 
of  the  second  be  taken  from  |-  of  the  first,  the  remainder  will 
be  6.  Ans.  48  and  36. 

21.  What  fraction  is  that,  if  5  be  added  to  its  numerator,  its 
value  is  2,  but  if  2  be  added  to  its  denominator,  its  value  is  ^  ? 

A71S.      |. 

22.  B  says  to  C,  if  3  years  were  taken  from  your  age  and 
added  to  mine,  I  should  be  twice  as  old  as  you.  C  replies,  if  3 
years  were  taken  from  your  age  and  added  to  mine,  our  ages 
would  be  the  same.     Required  their  ages. 

A?is.     B's  age  21,  C's  age  15  years. 

23.  It  is  required  to  find  two  numbers,  so  that  §  of  the  first ' 
added  to  f  of  the  second  shall  be  15|,  and  if  |  of  the  second  be 
subtracted  from  |-  of  the  first,  the  remainder  shall  be  5-f  |-. 

Am.     10  and  12. 

24.  It  is  required  to  divide  50  into  two  such  parts  that  f  of 
the  larger  shall  be  equal  to  f  of  the  smaller. 

Ans.     32  and  18. 

25.  A  gentleman,  at  the  time  of  his  marriage,  found  that  his 
wife's  age  was  to  his  as  3  to  4 ;  but,  after  they  had  been 
married  12  years,  her  age  was  to  his  as  5  to  6.  Bequired  their 
ages  at  the  time  of  their  marriage. 

A71S.     The  man's  age  24,  his  wife's  18  years. 

26.  A  farmer  hired  a  laborer  for  ten  days,  and  he  agreed  to 

pay  him  12  shillings  for  every  day  he  labored,  and  he  was  to 

forfeit  8  shillings  for  every  day  he  was  absent,  and  he  received 

at  the  end  of  his  time  40  shillings.     How  many  days  did  he 

labor,  and  how  many  days  was  he  absent  ? 

jhis.     He  labored  6  days,  and  was  absent  4. 
0# 


102  ALGEBRA. 

27.  A  gentleman  bought  a  horse  and  chaise  for  $208,  and  | 
of  the  cost  of  the  chaise  was  equal  to  f  the  price  of  the  horse. 
What  was  the  price  of  each  ? 

Ans.     Chaise,  $112;  horse,  $96. 

28.  A  and  B  engaged  in  trade,  A  with  $240,  and  B  with 
$96.  A  lost  twice  as  much  as  B ;  and,  upon  settling  their 
accounts,  it  appeared  that  A  had  three  times  as  much  remaining 
as  B.     How  much  did  each  lose  ? 

^-9/. -x  Ans.     A  lost  $96,  and  B  lost  $48. 

29.  Two  men,  A  and  B,  agree  to  dig  a  well  in  10  days,  but, 
having  labored  together  4  days,  B  agreed  to  finish  the  job, 
which  he  did  in  16  days.  How  long  would  it  have  required  A 
to  complete  the  labor  ?  -  -  Ans.     9§  days. 

30.  A  merchant  has  two  kinds  of  grain,  one  at  60  cents  per 
bushel,  and  the  other  at  90  cents  per  bushel,  of  which  he  wishes 
to  make  a  mixture  of  40  bushels  that  may  be  worth  80  cents 
per  bushel.     How  many  bushels  of  each  must  he  use  ? 

Am.     13^  bushels  of  60  cents,  26|  of  90  cents. 

31.  A  farmer  has  30  bushels  of  oats,  at  30  cents  per  bushel, 
and  which  he  would  mix  with  corn  at  70  cents  per  bushel,  and 
barley  at  90  cents  per  bushel,  so  that  the  whole  mixture  may 
consist  of  200  bushels,  at  80  cents  per  bushel.  How  many 
bushels  of  corn,  and  how  many  of  barley,  must  he  mix  with  the 
oats  ?  A71S.     10  bushels  of  corn,  and  160  of  barley. 

32.  A  drover  sold  6  of  his  oxen  and  8  of  his  cows,  and  he 
then  found  he  had  twice  as  many  oxen  as  cows.  But  after  he 
had  sold  10  more  of  his  oxen,  he  found  he  had  2  more  oxen 
than  cows.     How  many  had  he  of  each  at  first  ? 

Ans.     30  oxen  and  20  cows. 

y 

33.  Four  times  the  larger  of  two  numbers  is  equal  to  six 
times  the  less,  and  their  sum  is  15.     Required  the  numbers. 

A71S.     9  and  6. 

34.  A  and  B  can  perform  a  piece  of  work  in  6  days,  A  and  C 
in  8  days,  and  B  and  C  in  12  days.     In  what  time  would  each 


NEGATIVE     QUANTITIES.  103 

of  them  perform  the  work  alone,  and  how  long  would  it  take 
them  to  perform  the  work  together  ? 

A71S.  A  would  do  the  work  in  9|  days,  B  in  16  days,  C  in 
48  days,  A,  B  and  C  together,  in  5^  days. 

35.  A  gentleman  left  a  sum  of  money  to  be  divided  among 
his  four  sons,  so  that  the  share  of  the  oldest  was  ^  of  the  shares 
of  the  other  three,  the  share  of  the  second  4-  of  the  sum  of  the 
other  three,  and  the  share  of  the  third  \  of  the  sum  of  the 
other  three  ;  and  it  was  found  that  the  share  of  the  oldest 
exceeded  that  of  the  youngest  by  814.  What  was  the  whole 
sum,  and  what  was  the  share  of  each  person  ? 

A?is.  "\Miole  sum,  S1'20 ;  oldest  son's  share,  SIO ;  second  son's, 
S30  ;  third  son's,  S24;  youngest  son's,  S26. 


SECTION    XI. 
NEGATIVE    QrAXTITIES. 

Akt.  1.53.  The  student  will  sometimes  find  that,  on  account 
of  his  misconception  of  the  question,  he  has  added  a  quantity 
which  should  have  been  subtracted,  or  that  he  has  subti-acted  a 
quantity  which  should  have  been  added. 

This  may  be  illustrated  by  the  following 

EXAMPLES. 

1.  The  length  of  a  ceiiain  field  is  a,  and  its  breadth  is  b; 

how  much  must  be  added  to  its  breadth  that  its  contents  may 

be  m  ? 

Let  X  =  the  quantity  to  be  added  to  its  breadth. 

Then  b-\-x=ih.e  breadth. 

And  a{b-\-x)=m,  the  contents, 

ab-\-ax=m. 

m 
b-\-x=-. 
a 

m 

X=z 0. 

a 


104  A  L  G  t:  B  1{  A  . 

2.  Let  the  length  of  the  field  be  10  rods,  and  its  breadth  6 
rods ;  how  many  rods  must  be  added  to  its  breadth,  that  the 
contents  of  the  field  may  be  80  square  rods  ? 

Let  X  =  the  quantity  to  be  added  to  the  breadth.  Then,  by 
the  above  formula, 

x= b=~—Q==2  rods,  the  quantity  to  be  added. 

VERIFICATION. 


10x6+2=80  square  rods. 

3.  Let  the  length  of  the  field  be  10  rods,  the  breadth  8  rods; 
it  is  required  to  find  what  quantity  must  be  added  to  the  breadth 
that  the  contents  may  be  60  square  rods. 

By  the  formula, 

77Z  60 

x= Z»=:t7:— 8= — 2  rods. 

a  10 

We  perceive  by  the  above  that  it  is  — 2  rods  which  are  to  be 
added,  and  not  -\-2  rods;  but  we  add  quantities  together  in 
Algebra  by  simply  writing  them  one  after  the  other,  with  their 
respective  signs,  so  that  — 2  added  to  -|-8  becomes  8 — 2=6, 
the  answer,  which  is  the  same  as  subtracting  -\-2  from  -|-8. 
And,  in  general,  adding  a  minus  quantity  brings  the  same  result 
as  subtracting  a  plus  quantity  of  equal  value,  and  vice  versa. 

VERIFICATION. 


10  X^ — 2=60  square  rods.     Ans. 

4.  Suppose  the  field  to  be  10  rods  long  and  8  rods  wide, 
it  is  required  to  ascertain  how  much  must  be  subtracted  from  its 
width  that  its  contents  may  be  60  square  rods. 

To  subtract  a  minus  quantity  is  the  same  as  to  add  a  plus 
quantity.  If,  therefore,  we  change  the  sign  of  x  in  the  formula 
first  obtained,  x  will  then  express  how  much  is  to  be  subtracted. 

Thus,  — x= 3, 

a 

or,  x=0 =:b — :r^=2  rods. 

a  10 


NEGATIVE     QUANTITIES.  105 


VERIFICATION. 


10x^—2=60  square  rods. 

6.  If  the  field  were  10  rods  long  and  8  rods  wide,  how  many 
rods  must  be  taken  from  its  width  that  its  contents  may  be  100 
square  rods  ? 

By  the  formula, 

x=^b =  b — TTr= — ^  rods. 

a  10 

That  is,  — 2  is  to  be  subtracted  from  -{-8 ;  or,  as  we  perform 
subtraction  in  Algebra  by  changing  the  sign  of  the  subtrahend, 
and  thus  annexing  it  to  the  minuend,  we  have 
8— (— 2)=8+2=10; 
so  that,  in  general,  subtracting  a  minus  quantity  is  the  same  as 
adding  a  plus  quantity  of  equal  value. 

6.  John  Smith,  at  the  time  of  his  marriage,  was  50  years 
old,  and  his  wife  was  40.  When  will  his  age  be  twice  that  of 
his  wife  ? 

Let  a:=the  time. 

Then,  50-}-a;=2x40+^. 

50+:r=80+2^. 
And,  a:==50— 80=— 30  years. 

As  the  answer  is  — 30  years,  it  is  evident  that  he  is  not  myw 
twice  as  old  as  his  wife,  but  30  years  ago  his  age  was  twice 
hers. 

VERIFICATION. 


50-30=40—30x2. 

20=20. 

7.  J.  Jones  is  40  years  old,  his  wife  30.     AVhen  will  they 
both  be  of  the  same  age  ? 
Let  a:=the  time. 

Then,  40+  x=.Z^-\-x. 

40—30=  x—x. 
And,  10=0. 

The  answer  being  zero,  it  is  certain  they  never  will  be  of  the 
same  age,  but  that  one  will  always  be  10  years  older  than  the 
other. 


106  ALGEBRA. 

8.  What  fraction  is  such,  that  if  2  be  added  to  its  numerator 
its  value  is  -I,  or  if  2  be  added  to  its  denominator  its  value  is  ^  ? 

A71S. 


-12' 

9.  What  fraction  is  such,  that  if  7  be  added  to  the  numerator 

its  value  is  nothing,  but  if  2  be  added  to  its  denominator  its 

value  is  infinite  ?  .  — 7 

Atis.     — ^. 

10.  What  fraction  is  such,  that,  if  4  be  added  to  its  numer- 
ator its  value  is  nothing,  but  if  10  be  subtracted  from  its 
denominator  its  value  is  1  ?  —  i^. 

THE    COURIERS.  P 

1.  Two  couriers  set  out  at  the  same  time  from  A  and  C,  and 
travel  towards  each  other  until  thej  meet.  The  distance  from 
A  to  G  is  m  miles.  The  first  courier  travels  a  miles  per  hour, 
and  the  second  b  miles  per  hour.  How  far  from  A  and  C  will 
they  meet  ? 

A  B  C  D 


Let  us  suppose  them  to  meet  at  B. 

And  let  x  =  the  distance  A  B. 

And  y  =  the  distance  B  C. 

Then  x+y  =  A  G  =  m. 

As  the  first  travels  x  miles  at  the  rate  of  a  miles  per  hour,  to 
find  the  time  he  will  travel  this  distance,  we  say, 

X 

As  a  miles  :  x  miles  :  :  1  hour  :  -  =  the  time  the  first  cou- 

a 

rier  will  travel  the  distance  A  B. 

y 

And,  as  b  miles  :  y  miles  :  :  1  hour  :  ^  hours  =  the   time 
^  b 

the  second  courier  will  travel  the  distance  B  C. 

As  both  couriers  set  out  at  the  same  time,  and  arrive  at  the 

same  time  at  C, 

Therefore  -=7-. 

a     b 

And  x—^, 

0 


NEGATIVE     QUANTITIES.  107 

If  we  substitute  this  value  of  x  in  the  j&rst  equation,  we 

have 

ay 

And  'iy-\-by=bm. 

bm 


Hence 


y  = 


a-\-b 


Substituting  this  value  of  y  in  the  equation  x=-^,  we  have 
a       bm        abm         am 


b     a-\-b     ah-\-h'     a+b 

The  values  of  x  and  y  in  the  above  equation  are  both  posi- 
tive. Therefore,  whatever  value  we  may  assign  to  «,  b  and  m, 
it  will  answer  the  conditions  of  the  question. 

This  may  be  illustrated  by  the  following  question  : 

2.  Two  men,  A  and  B,  set  out  from  two  places,  distant  from 
each  other  144  miles,  and  travel  towards  each  other.  A  goes 
12  miles  an  hour,  and  B  four  miles  an  hour.  How  far  must 
each  travel  before  they  meet  ? 

By  the  above  formulae, 

x= — r-T=  -..->,.     =  108  miles,  the  distance  A  travels. 
a-\-b       12-f4 

bm      4x144 
And  7/  =  — --==-————==  36  miles,  the  distance  B  travels. 
-^        a-\-b      12-1-4 

VERIFICATION. 

108  +  36=144  miles. 

3.  If  the  couriers  were  to  set  out  at  the  same  time  from  A 
and  B,  and  travel  towards  C,  both  going  the  same  direction,  the 
first  going  a  miles  per  hour,  and  the  second  b  miles  per  hour, 
and  the  distance  A  B  being  m,  how  far  would  each  travel  before 
they  meet,  suppose  at  a  point  C  ? 

F A B C D 

Let  X  =  the  distance  A  C. 
And  y  =  the  distance  B  C. 
Then  x-y=AC—BC=AB=z7n. 


108  ALGEBRA. 

By  performing  the  same  operation  as  in  the  first  question, 
"w^e  find 

a     3' 

and  x=-f-, 
o 

Therefore         ~- — 7iz=m. 
b      -^ 

And  ay — ly=.hin. 

bm 


Whence  y-. 


a — b 


Substitute  this  last  value  of  y  in  the  former  equation,  and 
we  have 

ay     a      bin         dbm        am 
b       b     a — b     ab — l?     a — b 

Here  it  is  evident  that  the  values  of  x  and  y  will  not  be 
positive,  unless  a  be  greater  than  b  ;  or,  in  other  words,  unless 
the  courier  which  sets  out  from  A  travels  faster  than  the  one 
that  sets  out  from  B,  he  will  never  overtake  him. 

4.  Suppose  the  first  courier  to  travel  9  miles  per  hour,  and 
the  second  6  miles  per  hour,  and  the  distance  A  B  to  be  18 
miles,  and  it  was  required  to  find  how  far  each  would  travel 
before  the  one  overtook  the  other. 

Then  «=9,  ^=:6,  and  7?z=18. 

And,  by  the  first  formula; 

am      9x18     p  ^     -i        i      t  ,     r.  •  ii 

-=o4  miles,  the  distance  the  first  courier  woukl 


a—b      9—6 

travel.    " 


And,  by  the  second  formula, 

bm      6x18     „^      .,        ,       ,.  ,  ,  . 

2/= ;=-?T — 77-=oo  miles,  the  distance   the   second  courier 

a — b      9 — o 

would  travel. 

We  perceive,  by  the  above  operation,  that  the  point  C,  where 


NEGATIVE     QUANTITIES.  109 

the  couriers  meet,  is  54 — 36=18  miles  further  from  A  than  B 
is,  which  is  equal  to  the  distance  ?//. 

5.  Again,  let  a=Q,  ^=9,  and  7?z=18 ;  or,  suppose  the  first 
courier  sets  out  from  A  and  travels  6  miles  an  hour,  and  the 
second  sets  out  at  the  same  time  from  B  and  travels  in  the 
same  direction  towards  C  at  the  rate  of  9  miles  per  hour.  What 
distance  will  each  travel  before  thej  meet  ? 

Bj  the  first  formula, 

am      6x18  .        ,,     ^    ,,        \ 

=  —  rfo  miles,  the  first  travels. 


d—b      6—9 
By  the  second  formula, 

bm      9x18  C-.     M        1 

x= ,=-7^ — 7^=  —  &4  miles,  the  second  travels. 

a — 0      b — 9 

Here  the  values  of  x  and  y  are  both  negative.  Now,  how  shall 
we  interpret  this  result  ?  AYhat  is  the  meaning  of  the  negative 
sign,  in  this  case  ? 

To  understand  this,  we  must  observe  that  we  began  by  sup- 
posing the  parties  to  be  travelling  towards  C,  and  any  motion 
in  this  direction  would  have  been  indicated  in  this  example,  as 
it  has  been  in  the  preceding  examples,  by  the  sign  -f--  But, 
when  the  sign  -\-  is  taken  to  indicate  motion  in  one  direction, 
the  opposite  sign  —  must  indicate  motion  in  the  opposite  direction. 
Hence  the  minus  sign,  resulting  as  above,  indicates  that  the 
parties,  in  order  to  meet,  must  travel,  not  towards  C,  as  we  at 
first  supposed,  but  in  the  opposite  direction,  towards  F,  a  point 
36  miles  from  A,  and  54  miles  from  B,  where  they  will  meet. 

6.  Again,  let  fi^=6,  b=Q,  and  ??z=18;  or,  we  will  suppose 
the  couriers  both  to  start  at  the  same  time  from  A  and  B,  and 
both  to  travel  in  the  same  direction  towards  C,  and  both  trav- 
elling at  the  same  rate  of  6  miles  per  hour,  the  distance  A  B 
being  18  miles.  What  distance  will  each  travel  before  they 
meet  ? 

10 


> 


110  ALGEBRA. 

By  the  first  formula,  x=. r,  or  =-7r» 

•^  G^ — o        a — a       U 

_6X18_108 
°^  ^— "0=^"  "0"* 

By  tne  second  lormula,  ?/= -,  or  =-7r> 

*'  a — h        a — a      0 

6x18     108 

As  both  couriers  are  travelling  in  the  same  direction,  and  at 
the  same  rate,  it  is  certain  they  will  never  meet,  but  the  dis- 
tance between  them  will  continue  the  same. 

■«v.    mi       ^         1  •      '3;?7^         108 

154.  Thereiore,  the  expression  —t~  or  — tt-j  or  any  quantity 

with  zero  for  a  denominator,  is  the  symbol  for  infinity ;  for  it  is 
well  known  that  the  value  of  a  fraction  depends  on  the  number 
of  times  the  numerator  contains  the  denominator,  or  the  number 
of  times  the  denominator  may  be  taken  from  the  numerator, 
until  nothing  shall  remain. 

It  is  certain  that,  if  a  be  greater  than  3,  however  small  the 
difference,  the  couriers  will  eventually  meet ;  but,  if  the  differ- 
ence between  a  and  h  be  less  than  any  assignable  quantity,  then 
X  and  y  may  be  considered  infinite. 

Again,  let  a=.h^  and  7?2=0. 

am      «X0      0 


Then  x=i 

And  y 


a—b        0        0' 
Im      ^XO      0 


a—b        0    ~0' 


From  the  above  we  infer  that  x  and  y  are  equal,  and  that 
each  is  equal  to  the  other. 

Thus,  x=.x. 

This  is  an  identical  equation,  and  the  values  of  the  unknown 
quantities  cannot  be  known  by  it. 

And,  as  7?z=0,  it  is  evident,  that  as  both  couriers  start  from 
the  same  point,  and  travel  at  the  same  rate,  and  in  the  same 


NEGATIVE     QUANTITIES.  11.1 

direction,  they  will  always  be  together,  and  therefore  cannot 
meet. 

We  say,  therefore,  that  the  ^,  in  this  case,  is  an  expression 
of  an  Indeterminate  Quantity,  because  that  x  and  y  may  be  any 
quantities  whatever. 

But  it  is  not  true  that  the  expression  %  is  always  the  sign  of 
an  indeterminate  quantity. 

155.  In  fractions,  when  the  numerator  and  denominator  have 
a  common  factor,  and  which  in  some  cases  becomes  zero,  and 
makes  the  fraction  assume  the  form  of  §,  but  which,  without 
that  factor,  has  a  definite  value,  the  expression  is  not  inde- 
terminate. 

The  following  fractions  are  examples  of  this  kind : 

n(m — 71) 
Now,  if  77«=?7,  the  value  of  the  quantity  is  §. 

But,  on  examination,  we  perceive  that  both  the  numerator 
and  denominator  have  the  common  factor  m — n. 

Therefore,  by  dividing  both  terms  of  the  expression  by  m — n^ 

lTl(.lTi-\~7l\ 

it  becomes  — ^^ ,  which,  if  77z=7z,  is  equal  to  2w. 

n  ^ 

x—\ 

The  value  of  the  expression  -, 

if  we  divide  both  terms  by  x — 1,  is  1  ;  but,  if  a;=l,  the  value 

is  a 

Again,  let  a:= 


m — n 

Then,  if  ??z=7z,  the  value  of  a:=^. 

But,  if  we  divide  both  terms  by  the  common  factor  m — ?i,  its 
value  is  ')r^-\-mn-\-r^^  and  then,  on  the  supposition  that  m=ni 
its  value  will  be  3w^ 

INDETERMINATION. 

156.  In  investigating  the  theory  of  indetermination,  we  find 
many  curious  results  and  apparent  absurdities. 

This  will  appear  evident  by  investigating  the  following  prob- 
lems. 


112  ALGEBRA. 

1.  If  it  be  admitted  that  a=l  and  x=l,  it  may  be  shown 
that  1  is  2  and  2  is  nothing,  or  any  assignable  quantity. 
Let  a=x. 

Multiplying  both  terms  of  the  equation  by  x,  we  have 

ax=x\ 

Subtracting  a^  from  both  members, 

Q  o  O 

ax — a=x' — a. 

Resolving  both  terms  into  factors, 

a[x — a)  =  {x — a)[x-\-a). 
Dividing  by  x — a, 

a=x-\-a. 
Substituting  a  for  its  value  x, 

a=a-\-a. 

Dividing  both  terms  by  c, 

1=1-1-1=2. 

Again,  we  have  found  above  that 

x"^ — a^=:ax — d^. 

Dividing  both  terms  by  the  common  factor  x—a,  we  have 

ax — a^ 

x-j-a= . 

X — a 

Now,  as  X  and  a  by  the  supposition  are  each  equal  to  1,  we 

see  that 

ivi— r 

14-1=    ^^     ^, 
^  1—1 

And  2=g. 

Thus  it  appears  that  we  have  dearhj  proved  that  1  is  2, 
and  2  any  assignable  quantity.  Q.  E.  D. 

The  fallacy  is  this,  that  if  nothing  be  divided  by  nothing  the 
quotient  is  any  assignable  quantity. 

This  principle  may  be  further  illustrated  by  considering  the 
following  identical  equation. 

Let  16=16. 

Resolving  into  terms,  12-|-4=12-}-4. 

Transposing,  4—4=12—12. 

Resolving  second  term  into  factors,  4 — 4=3(4 — 4). 

Dividing  by  4 — 4,  1=3. 


NEGATIVE     QUANTITIES.  113 

Thus  it  appears  that  1  is  3 ;  and,  in  the  same  manner,  a  unit 
may  be  proved  to  be  any  definite  number. 

From  various  articles  in  the  foregoing  section,  we  infer  the 
following : 

1.  If  zero  be  multiplied  by  zero,  or  any  assignable  quantity, 
the  product  will  be  zero. 

2.  If  zero  be  divided  by  zero,  the  quotient  may  be  zero,  or 
any  assignable  quantity. 

3.  If  zero  be  divided  by  any  quantity,  the  quotient  will  be  zero. 

4.  If  any  quantity  be  divided  by  zero,  the  quotient  will  be 
infinity. 

5.  If  any  quantity  be  added  to  or  taken  from  infinity,  the 
result  will  be  infinity. 

6.  If  zero  be  multiplied  by  infinity,  the  product  may  be  any 
quantity. 

7.  K  infinity  be  divided  by  infinity,  the  quotient  may  be  any 
assignable  quantity. 

8.  One  infinity  may  be  infinitely  larger  than  another. 


SECTION   XII. 
Theorem  I. 

AuT.  157.  If  a  binomial  be  multiplied  into  itself,  the  pro- 
duct will  be  equal  to  the  sum  of  the  squares  of  both  terms,  plus 
twice  the  product  of  the  terms. 

XoTE.  —  The  theorems  in  the  following  section  may  be  illustrated  h^ 
diagrams,  and  it  would  be  well  for  the  pupils  to  draw  them. 

When  a  number  or  quantity  is  multiplied  into  itgelf,  the  product  is  a 
square. 

EXAMPLES. 

1.  Multiply  a-}-b  into  itself. 


ALGEBRA. 

VERIFICATION. 

8+4=12 

12 

8+4=12 

12 

64+32 

144 

32+16 

114 


a-\-b 

a^-\-ab 

ab^b'' 

d'J^^ab^b^  64+64+16=144. 

We  perceive,  by  the  above  operation,  that  the  square  of  any 
Hnomial  may  be  readily  obtained. 

2.  Multiply  Za-\-'lb  into  itself. 

3aX3a+2x3flX2^+23x23= 

9a^+12a3+43l 

3.  Multiply  x-\-2y  into  itself.  Ans.  x'+ixy-^-^y'. 

4.  Multiply  dab-\-?n  into  itself.  Atis. 

5.  Multiply  by-\-4:X  into  itself.  Aiis. 

6.  Multiply  2m-\-o?i  into  itself.  A7is. 

7.  Multiply  7fZ+2e  into  itself.  A?is. 

8.  Multiply  2?^+3^^?  into  itself.  Arts. 

9.  Multiply  5«-+23  into  itself.  Ajis. 

10.  Multiply  1+^  into  itself.  A?is. 

11.  Multiply  3+^  into  itself.  A?is. 

12.  Multiply  2+J-  into  itself.  A?is. 

Theorem  II. 

158.  If  the  sum  of  two  numbers  or  quantities  be  multiplied 
by  their  difference,  the  product  will  be  equal  to  the  difference 
of  their  squares. 


Multiply  a-{-b  into  a — b. 

a+b 

a—b  8—4=  4  4 


VERIFICATION. 

a-\-b  .  8+4=12  12 


ci'-^ab  64+32  48 

-ab—b''  —32-16 

cr        -b""  64         -16=48. 


NEGATIVE    QUANTITIES.  115 

Theorem  III. 

159.  If  the  difference  of  two  numbers  or  quantities  be  mul- 
tiplied into  itself,  the  product  will  be  equal  to  the  sum  of  their 
squares,  minus  twice  their  product. 


EXAMPLES. 

.  Multiply  a- 

a — b 
a — b 

-b  into  a- 

-b. 

VERIFICATION. 

12-8=  9 
12-3=  9 

9 
9 

a^ — ab 
-ab-\-b'' 

144-36 
-36-f9 

81 

a^—2ab-\-b\  144—72+9=81. 

2.  Multiply  2>a—2b  into  3a— 23.         Ans.  ^d'—12ab-\-^b\ 

3.  Multiply  bm — ?i  into  bm — ?^. 

4.  Multiply  Aab—x  into  ^ab—x. 

5.  Multiply  ^a'—b^  into  ^d'—b^ 

6.  Multiply  a;* — y^  into  x'^ — if. 

Note. — If  the  square  of  the  difference  of  two  numbers  be  subtracted 
from  the  square  of  their  sum,  the  remainder  will  be  equal  to  four  times 
their  product. 

Thus,  {a-\-hf-{a-bf={dJ^2ab^b'')  —  [d—2ab-\-b'')  =4.ab. 

Theorem  IV. 

160.  If  twice  the  product  of  two  quantities  be  subtracted  from 
the  sum  of  their  squares,  the  remainder  will  be  equal  to  the 
square  of  their  difference. 

(^a^^P)—2ab=ar—2ab-\-b\ 

But  this  expression,  by  Problem  3d,  is  the  square  of  their 
difference. 

VERIFICATION. 

Let  9  and  3  be  the  two  numbers. 

Then  (92+32)-(2x9X3)=(9— 3)^ 

(81+9)-(54)=36. 

90-54=36. 

36=36. 


116  ALGEBRA. 

Theorem  Y. 

161 1  If  there  be  two  quantities,  one  of  which  is  divided  into 
any  number  of  parts,  the  product  of  the  two  quantities  will  be 
equal  to  the  product  of  the  undivided  number  into  the  several 
parts  of  the  divided  number. 

Let  the  two  quantities  be  a  and  3,  and  let  h  be  divided  into 
three  parts,  c,  c?,  and  e. 

Then  3=c+^+e. 

And  ab^=^ac-\-ad-\-ae, 

VERIFICATION. 

Let  the  two  numbers  be  12  and  10,  and  let  10  be  divided 
into  the  parts  5,  3,  and  2. 

Then  10=5+3+2. 

And  12X10=12X5+12><3+12X2: 

120=60+36+24. 
120=120. 

Theorem  VI. 

162.  If  any  quantity  be  divided  into  two  parts,  the  square  of 
this  quantity  will  be  equal  to  the  sum  of  the  products  of  this 
quantity  into  its  two  parts. 

Let  a  represent  the  quantity,  and  h  and  c  the  parts  into 
which  it  is  divided. 

Then  «=:3+c. 

And  aX«^=«^(^+c). 

VERIFICATION. 

Let  12  be  divided  into  two  parts,  9  and  3. 

Then  12=9+3. 

12x12=12(9+3). 
144=108+36=144. 

Theorem  VII. 

163.  If  any  quantity  or  number  be  divided  into  two  parts, 
the  product  of  the  whole  and  one  of  the  parts  will  be  equal  to 


THEOREMS.  117 

the  product  of  the  two  parts,  plus  the  square  of  the  aforesaid 
part. 

Let  a  represent  the  whole  quantity,  and  h  and  c  the  parts. 

Then  a^=^h-\-c. 

Multiplying  both  sides  of  the  equation  by  3,  we  have 

YEEIFICATION. 

Let  12  represent  the  number,  and  9  and  3  the  parts  into 
which  it  is  divided. 
Then  12=9+3. 

Multiplying  both  parts  of  the  equation  by  9,  we  have 
9x12=9(9+3). 
108=81+27=108. 

Theorem  A^III. 

164i  If  any  quantity  be  divided  into  two  parts,  the  square  of 
the  whole  quantity  will  be  equal  to  the  squares  of  the  two  parts, 
plus  twice  their  product. 

Let  a  represent  any  quantity,  and  h  and  c  the  parts  into 
which  it  is  divided. 

Then  a=.h-\-c. 

By  squaring  both  sides  of  the  equation,  we  have 

VERIFICATION. 

Let  9  be  divided  into  two  parts,  6  and  3. 

Then  9=6+3. 

By  squaring  both  parts  of  the  equation,  we  have 

9-=(6+3)l 

81=36+3G+9=8L 

Theorem  IX. 

165.  If  any  number  or  quantity  be  divided  into  two  equal 
parts,  and  into  two  unequal  parts,  the  square  of  one  of  the 
equal  parts  will  be  equal  to  the  product  of  the  two  unequal 


118  ALGEBRA. 

parts,  plus  tlie  square  of  half  the  difference  of  the  two  unequal 

parts. 

Let  a  represent  one  of  the  equal  parts,  and  h  and  c  the  two 
unequal  parts. 

Then  CL=-^' 

And  2a=3-]-c. 

We  now  add  — 43c  to  both  sides  of  the  equation. 
And  4«2_43c=— 43c4-3'+2ic+cl 


d^—  bc=. 


4 

52_23c+c^ 


4 

VERIFICATION. 

Let  12  be  divided  into  two  equal  parts,  6  and  6 ;  and  into 
two  unequal  parts,  9  and  3. 

And  36=27+?14i±^. 

'  4 

36=27+9=36. 

Theorem  X. 

166.  If  any  quantity,  2a,  be  divided  into  two  equal  parts,  and 
if  any  quantity,  b,  be  added  to  2<2,  the  product  of  2a-{-b  into  />, 
plus  the  square  of  a,  will  be  equal  to  the  square  of  a-\-b.  Then, 
by  the  proposition,  2a-\-b  will  be  the  whole  quantity. 

Multiplying  by  b,  we  have 

b{2a-\-b)=2ab+b\ 
By  adding  a^  to  each  member  of  the  equation,  we  have 

b{2a+b)+a'=a'+2ab-{-b\ 
Therefore  b{2a+b)-\-d'={a+b)\ 


THEOREMS.  119 


VERIFICATION. 

Let  a=10,  and  b=2. 


Then  2(2xl0  +  2)+10'^=(10+2f ; 

144  =    144 

Theorem  XI. 

167.  If  any  quantity  be  divided  into  two  parts,  the  square  of 
this  quantity,  and  one  of  the  parts,  will  be  equal  to  twice  the 
product  of  the  whole  quantity  and  that  part,  plus  the  square  of 
the  other  part. 

Let  the  whole  quantity  be  denoted  by  a,  and  the  parts  by  b 
and  c. 

Then  a=b-\-c. 

And  a — c=b. 

d'-\-c'=2ac-\-b\ 

VERIFICATION. 

Let  12=3+9. 


Then  12'^+9^=2xl2x9  +  3^ 

And  144+81=216+9. 

225=225. 

Theorem  XII. 

168.  If  any  quantity  be  divided  into  any  two  parts,  four 
times  the  product  of  the  whole  quantity  into  one  of  the  parts, 
plus  the  square  of  the  other  part,  will  be  equal  to  the  square  of 
the  quantity  which  consists  of  the  whole  and  the  first-mentioned 
part. 

Let  a  represent  the  quantity,  and  b  and  c  the  two  parts  into 
which  it  is  divided. 

Then  a=i+c. 

Multiplying  both  members  of  the  equation  by  Ab,  we  shall 
have  43xa=43X(^+c). 

4a5=4i^+45c. 


120  ALGEBRA. 

We  now  add  &  to  both  members. 
Or  ^ab-^e={c^nf. 

VERIFICATION. 

Let  a=12,  and  b=9,  and  c=3. 

Then  12=9+3. 

And  4x12x9+3^=  (3+2x9)'^.    /  ? _  ^  t  - '^  ^j 

441         =       441  ^^ 

Theorem  XIII. 

169.  If  any  quantity  be  divided  into  two  equal  parts,  and  also 
into  two  unequal  parts,  the  sum  of  the  squares  of  the  two 
unequal  parts  will  be  double  the  square  of  half  the  quantity,  plus 
twice  the  square  of  the  quantity  which  consists  of  the  diiference 
oetween  half  the  quantity  and  the  larger  of  the  unequal  parts  of 
the  quantity. 

Let  2a  represent  the  quantity,  and  a  =  one  of  the  equal 
parts,  and  b  =  half  the  difference  between  the  equal  and  un- 
equal parts. 

Then  a-\-b  =:  the  larger  part. 

And  a — b  =  the  less  part. 

And  [a-\-by'-{-{a — by  =  the  sum  of  their  squares. 

But  (d'-{-2ab-\'b'')-i-[ar—2ab+b')=2a'-\-2b\ 

And  2a'^-\-2b^  =  twice  the  square  of  half  the  quantity,  plus 
twice  the  square  of  half  the  difference  between  the  equal  and 
unequal  parts  ;  that  is,  the  difference  between  half  the  quantity 
and  the  larger  of  the  unequal  parts. 

VERinCATION. 

Let  10=7-1-3;  10-^2=5;  7—5=2. 


Then  (5-}-2)^+(5-2)'=2x5'+2x2^ 

And  49  +  9=50+8. 

58=58. 


INVOLUTION.  121 

Theorem  XIV. 

170.  If  any  quantity,  2a,  be  divided  into  two  equal  quantities, 
a  and  a  ;  and,  if  any  quantity,  3,  be  added  to  2«,  the  square  of 
2a+^,  plus  the  square  of  ^,  will  be  equal  to  twice  the  square  of 
(2,  plus  twice  the  square  of  «+^- 

Now       (2a+3)2+3-=4a^+4a^+3^+^'=4a-+4a3+2i\ 
But  4c-+4a^4-2^»'=2a-+2(fl+i)-l 

Therefore  (2a+3)^4-3==2a^+2(a+3)l 

VERIFICATION. 

Let  G=10;  and  3=4. 

Then        (2x10+4)^+4^=2(10'-) -1-2(10+4)1 
And  576+16=200+392. 

Therefore  592=592. 


SECTION   XIII. 

INVOLUTION. 

AuT.  171.  Involution  is  the  raising  of  powers  from  any  pro- 
posed root ;  or,  the  method  of  finding  the  square,  cube,  biquad- 
rate,  &c.,  of  any  given  quantity. 

172.  A.  pmver  is  the  product  of  any  quantity  multiplied  into 
itself  a  certain  number  of  times,  and  the  degree  of  the  power  is 
denoted  by  an  exponent  written  over  the  root.  Thus  a^  is  the 
third  power  of  a,  and  a  is  the  root. 

173.  The  exponent^  or  index.,  shows  how  many  times  the  root 
has  been  used  as  a  factor. 

Thus,  G^X^X«X«=^S  and  a:X^=^^- 

174.  When  a  quantity  is  written  without  any  index,  its  index 
is  uniformly  considered  a  unit.   Thus,  a=a^,  and  x=x^.     There- 

11 


122  ALGEBRA. 

fore,  to  raise  any  quantity  to  any  required  power,  the  pupil  will 
see  the  propriety  of  the  following 

Rule.  Multiply  the  index  of  the  quardity  by  the  index  of 
the  power  to  which  it  is  to  be  raised ^  and  the  result  will  be  the 
paiver  required. 

Or,  multiply  the  quantity  into  itself  as  many  times,  less  one. 
as  is  denoted  by  the  index  of  the  power,  and  the  last  product  will 
be  the  answer. 

175.  When  the  sign  of  any  simple  quantity  is  -|-,  all  the 
powers  of  it  will  be  +  ;  and  when  the  sign  is  — ,  all  the  even 
powers  will  be  -}-,  and  the  odd  powers  — ,  as  is  evident  from 
multiplication. 

EXAMPLES. 

1.  What  is  the  fifth  power  of  a.?  Ans.  a^. 

2.  What  is  the  third  power  of  ax  ?  Ans.  a^x^. 

3.  Required  the  square  of  a^x.  Atis.  c^t^. 

4.  Required  the  cube  of  — 3a^  Aiu.  — 27fl^^ 

5.  Required  the  fourth  power  of  —aV^&.  Ans.  a%^c^^. 

2ax^  4a~x^ 

6.  Required  the  square  of  — ^7—.  Ans.  . 

7.  Required  the  fifth  power  of  2ab''x\  A7is.  S2a'b'°x''. 

8.  Required  the  sixth  power  of  f  a^z^.  Ans.  y^/^a'lr'^. 

9.  Required  the  third  power  of  2a~^.  Ans.  S^r^. 

10.  Required  the  fourth  power  of  —  3m~^.         Ans.  81m~-^ 

11.  Required  the  ?7ith  power  of  a".  A7is.  a'"". 

12.  Required  the  fourth  power  of  2a;"'.  Ans.  I62:'''". 

,  Sa'b^  ^         27a'b' 

13.  Required  the  third  power  of  -j — 5-.  Ans»    „     3  ,^. 

176.  Polynomials  are  involved  by  multiplying  the  quantity 
by  itself  as  many  times,  ivanting  one,  as  there  are  units  in  the 
exponent  of  the  power. 


INVOLUTION. 


123 


14.  Let  a-}-b  be  raised  to  the  fifth  power. 
a-\-b 


a^-\-ab 
-{-ab-^-b"- 

^aJ^lYz=za^J^1ab^V' 
a-\-b 


a?J^2a%-\-ab'' 
J^a^b^2ab''-^P 

{a+bf=za?-]-Zd'b-\-Zab''+b^ 
a-\-b 


a'^^a?b-]-^a%''-\-ab^ 
\.a^b-]-^d'b''-\-^aW-\-b^ 

(^a-\-by=a'-\-4:a?b-\-U''b''-\-^ab^-\-b' 
a-{-b 

a'j^4.a'b^QaW-\-4:a%^-^ab' 
j^a'b-\-^a%^-\-Qa%^^^ab'+b' 

{a-\-bY=a'-\-^a'b^l^a?b''-\-10a%^-\-bab'+b^ 

Kequired  the  third  power  of  a — b. 

(^a—bf=a—b 
a — b 


d^ — ab^ 
—ab+b'' 

{a^bf=a^—2ab+b^ 
a — b 


a^—2a'b-{-ab'' 
—a'b+2ab''—b^ 


{a—bY=a?—U%+Zah^-b^ 


1st  power. 


2d  power. 


3d  power. 


4th  power. 


5th  power. 
1st  power. 


2d  power. 


8d  power. 


15.  Required  the  fifth  power  of  x — 2y. 

A71S.  x'—10x'y-^^0xy—S0xy-\-S0xy'S2y\ 


124  ALGEBRA. 

16.  Required  the  third  power  of  a — b-\-l. 

17.  Required  the  second  power  of  2x'^ — Sx-{-4. 

18.  Required  the  sixth  power  ofx — 2. 

Ans.  x'—12x'+QQx'—lQ0x''-{-240x''—ld2x+G4:, 

2x-7j-^ 

19.  Required  the  second  power  of  . 

Ans.  ^ 


W—2\bd^\U' 

20.  Required  the  fourth  power  of  «'"—«". 

Ans.  tt^"'—4a^'"+"-f6«""'-^'^"—4a"'+''" +«■'". 

21.  What  is  the  second  power  of  2x^—ox-\-^  ? 

Ans.  4x^—\2o?-^\\x'—Zx-\-\. 

22.  What  is  the  third  power  of  a-|-23— c  ? 

Ans.  a^-\-^a^b—^a\-^\2ay'—\2abc^Za(?^W—\2b''c^Uc^ 
-c\ 

23.  What  is  the  fourth  power  of  a-\-b-\-c-\-dl 

Ans.  a'-\-4a%^^a%''-^4ab^-^b'-\-4a\-\A2a%c-\-\2ab\-\-4b^c 
■\-\aH^Vla%d-Y\-2abH-^Wd^U^c~^\2ab&  +  We  -f  \2a^cd 
■\-24abcd-\-\2b\d  4-  6aV  _|_  i2a^^2  _|_  532^2  _^  4^^3  _^  i2ac'^  -f- 

\2acd^-\-4ad'^-\-4b&-\-\2bc'd^\2bcd^-\-4bd^^c'-^4&d  -f  6c*'d^^  + 
4c^^+^^ 

24.  What  is  the  second  power  of  x^-\-27?-\-x-\-2  ? 

Ans.  a;«+4a;^+6:?;^+8:c^+9:c2_}_4^_|_4^ 

a     b  a^  b'^ 

25.  What  is  the  second  power  of- — -?     Ans.     —^ — 24--. 

u     a  0  a 

26.  What  is  the  third  power  oix} — x — 1  ? 

Ans.  x^—Zx^-\-hx?—Zx—\, 

27.  What  is  the  third  power  of  a — b — 2c' — d^  ? 

Ans.  a^— 3a'A+3a32— ^3— 6aV+12aJc^— 6^V— 3a'^3+6aW' 
-Z})'d^-\-Vlac'-\-\2aed^^Zad'—Vlbo'  —  Ubc'd'-  Ud'  —  8c'  — 
l2c'd^—Qed'—d\ 


EVOLUTION.  125 


SECTION   XIV. 

EVOLUTION,    OR  THE   EXTRACTION   OF   ROOTS. 

Art.  177 1  Evolution  is  the  reverse  of  involution,  being  the 
method  of  finding  the  roots  of  any  given  quantity.  It  will, 
therefore,  be  necessary  to  trace  back  the  steps  of  the  operation 
in  involution. 

Hence,  to  find  any  root  of  a  monomial,  we  adopt  the  fol- 
lowing 

Rule.  Extract  the  required  root  of  the  coefficient  for  the 
coefficient  of  the  answer^  and  the  root  of  the  quantity  subjoined 
for  the  literal  part  of  the  answer. 

178i  If  the  quantity  proposed  be  a  fraction,  its  root  will 
be  found  by  taking  the  root  both  of  its  numerator  and  denom- 
inator. 

179.  The  square  root,  the  fourth  root,  or  any  other  even  root 
of  an  affirmative  quantity,  may  be  either  plus  or  minus. 

Thus,  A/a^=+a  or  — a  ;  and  /s/b^=-\-b,  or  — b.  But  the 
cube  root,  or  any  other  odd  root  of  a  quantity,  will  have  the 

same  sign  as  the  quantity  itself.     Thus  /^a^=a ;  /^ — a^=—a^ 

and  /s/ — a^-= — a. 

The  reason  why  -\-a  and  — a  are  each  the  square  root  of  a*^, 
is  obvious;  since,  by  the  rule  of  multiplication,  (+a)X(+«) 
and  ( — a)X( — ^)  are  each  equal  to  cr. 

180.  In  the  case  of  the  cube  root,  fifth  root,  &c.,  of  a  nega- 
tive quantity,  the  rule  is  equally  plain ;  since,  by  multiplying, 
we  have  (•— o)X(— a)X( — a)  =  —a^. 

It  may  also  be  stated  here  that  any  even  root  of  a  negative 
quantity  is  unassignable ;  or,  as  it  is  usually  called,  imaginary. 

Thus,  a/ — (^  cannot  be  determined,  as  there  is  no  quantity, 
either  positive  or  negative,  that,  when  multiplied  by  itself,  will 
produce  — d^. 

11^ 


126  ALGEBRA. 


EXAMPLES. 


1.  Find  the  square  root  of  9a^ 

Here  A/Va^=A/^X/s/a^=^Xci=Ba.     Am, 

2.  "What  is  the  cube  root  of  8a;^? 

Here  /^'^=/^X/^x^=^Xx=2x.    Ans. 

3.  It  is  required  to  find  the  square  root  of  -^-. 


Here 


a^b'^     fj  c^lp-     ah 

— — -L — Au'i 

^a'b' 


4.  What  is  the  cube  root  of      ^^„  3 


3 


Here  — 


Sa'b'  /^X^<r^         ^Xab-"         2ab' 


Atis. 


27c^          /^'llx^C           3Xc  3c 

5.  What  is  the  square  root  of  16a^5^?  Ans.  Ad^b'^. 

6.  What  is  the  cube  root  of  — l^bx^'if  ?  Atis.  — bx'ip. 

7.  What  is  the  fourth  root  of  %la'b^  ?  Ans.  Ub\ 

o    11m      .     1     ^^,             r.  32mV\  ,        2mn^ 

8.  What  IS  the  fifth  root  of               ?  Ans.   — - — . 

9.  What  IS  the  sixth  root  of    ,^^^    ?  Ans.  —r-, 

4096  4 

Note.  —  Fractions  sliould  first  be  reduced  to  their  lowest  terms. 

10.  Required  the  square  root  of  ^       ^.  Ans.   -^. 


EVOLUTION    OP    POLYNOMIALS. 

181 1  To  extract  the  square  root. 

Since  the  square  of  a-{-b  is  a^-\-2ab-{-b^y  in  order  to  obtain 
the  square  root  of  a'^-^2ab-\-b'^,  we  must  consider  by  what  pro- 
cess the  quantity  a-{-b  can  be  generally  derived  from  it. 

Now,  in  the  first  place,  we  observe  that  a,  the  first  term  of 
the  root,  is  the  square  root  of  a^,  the  first  term  of  the  square  * 


EVOLUTION.  127 

and,  in  addition  to  this,  there  still  remains  2ab-\-b'\  from  which 
b  is  to  be  obtained;  but  2«^-{-^^  is  the  same  as  {2a-\-b)b ;  and, 
therefore,  b  will  be  determined  by  dividing  the  first  term  of  the 
remainder  by  twice  the  first  term  of  the  root.  To  complete 
the  operation,  twice  this  first  term,  together  with  the  second, 
must  be  multiplied  by  the  second ;  and,  after  subtraction,  there 
is  no  remainder. 

182.  If  the  proposed  quantity  consists  of  more  terms,  it  is 
evident  that  we  have  only  to  consider  a-\-b  in  the  place  of  a, 
and  then,  by  the  same  process,  another  term  of  the  root  will  be 
obtained,  and  so  on ;  and  hence  we  have  the  following 

GrENERAL  RuLE.  Arrange  the  terms  in  the  order  of  the  mag- 
nitudes of  the  indices  of  some  one  quantity. 

Find  the  square  root  of  the  first  term,  and  subtract  its  square 
from  the  proposed  quantity. 

Bring  down  the  next  tivo  terms,  and  find  the  next  term  of  the 
root  by  dividing  this  last  quantity  by  twice  the  first,  and  affix  it, 
with  the  proper  sig7i,  to  the  divisor. 

Multiply  this  result  by  the  second  term  of  the  root,  and  bring 
down  to  the  remainder  as  many  terms  as  make  the  number 
equal  to  that  of  the  next  completed  divisor  ;  and  thus  continue 
the  process,  till  the  root,  or  the  requisite  appi'oximation  to  it,  be 
obtained. 

See  National  Ahithmetic,  page  243. 

EXAMPLES. 

1.  Find  the  square  root  of  x^ — ^x?y'^-\-^y^, 

x'^-~Qxhf-\-^y\x^-^f 
x' 


— 6:cy4-V. 

183.  If  the  terms  had  been  arranged  in  the  reverse  order, 
as  9z/^ — 62;y-{-a;^  the  root  would  have  been  found  by  a  similar 
process  to  be  Zy^ — x^,  which  difi"ers  in  its  sig7i  from  the  former. 


128  ALGEBRA. 

The  reason  of  this  is,  that  the  square  root  of  a  quantity  may 
be  either  positive  or  negative,  agreeably  to  Art.  179 ;  and  in 
the  first  case  we  have  one  sign,  in  the  second  the  opposite. 

2.  Find  the  square  root  o^  4x^—4:X^—Sx^-{-2x-{-l, 

4x^—^x'—Sx''-\-2x+l{2x''—x—l. 
4x' 


—4:X^-{-    X"^ 


4x'—2x-l)—4:x''-{-2x-i-l 
—4x^-\-2x-}-l 

3.  Extract  the  square  root  of  16  (a^+1)— 24a(a'+l)-}-41al 

Having  arranged  the  terms  according  to  the  dimensions  of  aj, 
we  have 

16a^-24as+41a'^-24a+16(4a^— 3a+4. 


Sa''-Sa)-24a'+4:la' 


8a2-6a+4)32a'^-24a+16 
32a--24a+16 


4.  Required  the  square  root  of 

4a^— 16a^2:^+16aM+20a%V— 40a*:c%^c^+25c2/^. 

4a'-lQJx^-{-lQjx^-{-20jy%'^—4:0Jx%'^c^+2^cyK 
4a'  (2j—4a^x^-\-5yh^. 

5  3.    2\  9.    2  2 

4:a^—4a^x-')—lQa^x'''-\-lQa- 


a  4 

X" 


9    2.  3     4 

—16a^x^-\-lQa-'x^ 


2.  32  51\  3     5     1  3251  J 

4a--'  —  ^aJx'^-{-57/'c^)20a^y^c^—40a^x^y'^c^-{-2bcy^ 


20Ji/c^—40a^x%jh^-{-2^cy^. 


EVOLUTION.  129 

5.  Extract  the  square  root  of  a^-^x^. 

,        /    ,  x^      x'  ,    x'        bx"     p 


a 


Ad' 


^'''^a      8aO     Ad' 


x^      x" 


2a -^      " 


Ad'     U'  '  64a« 


a     4a^  '  16a^  )  ^a>     64a« 


8a^  '  16a«      64a« 


■^a     4a^"^8^     128^7     64^«"^"64^ 

bx^       bx'' 
64^«     128^ 


,  &c. 


6.  What  is  the  square  root  oi  x'—2x^^Zx'—2x-\-l  ? 

Ans.  a? — x-\-\, 

7.  What  is  the  square  root  Q'ix^—1x^^x''-{-l7?—1x^^\  ? 

Alts,  y? — o?-\-\. 

8.  What  is  the  square  root  of  a'-^Aa%-^\^a%''^\1db^^W  ? 

Ans.  d'^2ab-\-m. 

9.  Extract  the  square  root  of  a* — 2a?-\-2d' — ci-\-^. 

Ans.  a' — a+2"- 

10.  What  is  the  square  root  of  4aV — \1(]^x^-\-\^o}x'—^(V'x 
-fa"?  Ans.  2ax''—Sa''x+a\ 

11.  What  IS  the  square  root  01  -r- — ■qTT'Q'a  • 

a    2b 


130  ALGEBRA. 

EVOLUTION   BY  DETACHED   COEFFICIENTS. 

1.  What  is  the  square  root  of  4:X^—4:X^J^lSx'^—6x-{-91 

4—4+13—6+9(2—1-1-3= 
4  2x''—x-{-d. 


4_1)_-4_|_13 
-4+  1 


4-2+3)12—6+9 
12-6+9. 


2.  What  is  the  square  root  of  9a;'— 242:*+12a;'+16;r^— 16a: 
+4? 

9+0-24+12+16—16+4(3+0-4+2= 

9  3a;3+0a;2— 4a;+2= 

6+0— 4)+0— 24+12+16  Sx^—4:X-}-2. 

_24—  0+16 

6+0—8+2)12+  0—16+4 
12+  0—16+4. 


3.  What  is  the  square  root  of  4a;«— 4a;5+12a;'+a;'— 6a;+9  ? 

4_|_0+0— 4+12+0+1— 6+9(2+0+0-1+3= 
4  2a:^+0a;3+0a;2— a;+3= 

4+0+0-l)+0  +  0-4+12+0+l        22;^— a;+3. 
_|-0+0— 4—  0—0+1 

4+0+0-2+3)12+0+0-6+9 
12+0+0-6+9 


The  pupil  will  perceive  that  the  5th  power  of  x  in  the  second 
question,  and  the  3d,  6th  and  7th  power  of  x  in  the  third 
question,  are  wanting;  therefore  their  place  in  the  operation 
must  be  supplied  by  zero. 

4.  What  is  the  square  root  of  4a^-16fi^'+24a'^— 16a+4? 

Ans.  2^2— 4a+2. 


EVOLUTION.  131 

5.  What  is  the  square  root  of  4x'''—12x^—12x^-\-9x''-]-lSz 
_|_9?  Ans.  2a;^— 3x— 3. 

6.  "What  is  the  square  root  of  16x'-\-2Ax'-\-^9x''-{-Q0x-]-100  ? 

Am.  4x'^-j-3a:4-10. 

7.  What  is  the  square  root  of  9a;'— 12a;^+10:c^— 28:c^-f  17x^ 
—  8:c-|-16?  Ans.  ox^—2x^-{-x—4. 

2      1 

8.  What  is  the  square  root  of  771^^+2771— 1 1 — rj 

Ans.   m-\-l . 

m 

EXTRACTION  OF  THE  SQUARE  ROOT  OF  NUMBERS. 

184.  As  numbers  are  not  expressed  in  the  same  manner  as 
algebraic  quantities,  it  is  evident  that  the  same  rule  for  ex- 
tracting the  square  root  of  algebraic  quantities  will  not  apply 
to  extracting  the  roots  of  numbers  without  additional  con- 
siderations. But,  if  the  foregoing  rule  be  assisted  by  the 
"  Method  of  Pointing,"  it  will  enable  us  to  extract  the  square 
root  of  numbers. 

185.  Since  the  square  root  of  1  is  1 ; 

the  square  root  of  100  is  10 ; 

the  square  root  of  10000  is  100  ; 

the  square  root  of  1000000  is  1000,  &c., 
it  is  evident  that  the  square  root  of  a  number  of  figures  less 
than  three  must  consist  of  only  one  figure ;  that  of  a  number 
more  than  two  figures  and  less  than  five,  of  two  figures ;  that 
of  a  number  more  than  four  figures  and  less  than  seven,  of  three 
figures,  and  so  on.  Whence  it  follows,  that,  if  a  dot  be  placed 
over  every  alternate  figure,  beginning  at  the  unit's  place,  the 
number  of  such  points  will  be  the  same  as  the  number  of  figures 
in  the  root. 

The  same  rule  may  be  extended  to  decimals,  by  first  making 
the  number  of  decimal  places  even,  and  then  commencing  at  the 
unit's  place  and  pointing  towards  the  right  hand  over  every 
alternate  figure,  as  before ;  and  the  number  of  such  points  will 
be  the  same  as  the  number  of  decimal  places  in  the  root. 


132  ALGEBRA. 

EXAMPLES. 

1.  Extract  the  square  root  of  273529. 

ARIl'HMETICAL  FORM.  SYMBOLICAL   FORM. 

273529(523  273529  (500-|-20+3 

25  500^=250000 


102)235  2X500+20=1020)23529 

204  20400 


1043)3129  2X(5004-20)+3=1043)3129 

3129.  3129 


The  pupil  will  perceive  that  both  these  operations  are  per- 
formed by  Art.  182. 

2.  Extract  the  square  root  of  45796.  Ans.  214. 

3.  Extract  the  square  root  of  106929.  Ans.  327. 

4.  Extract  the  square  root  of  36372961.  Am.  6031. 

5.  Extract  the  square  root  of  22071204.  Am.  4698. 

6.  Extract  the  square  root  of  33.1776.  Ans.  5.76. 

7.  Extract  the  square  root  of  .9409.  An^.  .97. 

8.  Extract  the  square  root  of  .0029997529.  Am.  .05477. 

9.  Extract  the  square  root  of  .001234.     Am.  .035128+. 

10.  Extract  the  square  root  of  32176552.863844. 

Am.  5672.438. 

CUBE   ROOT. 

ISO,  Investigation  of  a  rule  for  extracting  the  Cube  Root  of 
a  compound  algebraical  quantity. 

Since  {a-\-bf=a?-\-Za'b-\-''6ab'^-^h^,  we  must  have  the  cube 
root  of  the  latter  quantity  =  o,-\-b  ;  and  our  object  is  to  deter- 
mine how  it  may  be  deduced  from  it. 

Now,  the  first  term  a  of  the  root  is  the  cube  root  of  a?,  and 
the  first  term  of  the  proposed  quantity ;  hence,  taking  away 
a^  we  have  2>a^b -\-Zab'^ -\-h^  left  to  enable  us  to  find  b;  but 
Za'b-^'6ah'-\-b^=^{^a'-^oab-]-y)b.  It  is,  therefore,  manifest 
that  h  wiU  be  obtained  by  dividing  the  first  term  of  the  re- 
mainder by  three  times  the  square  of  a;  and,  to  complete  the 


EVOLUTION.  133 

divisor,  we  must  add  to  Sa^  three  times  the  product  of  the  two 
terms,  or  Sab,  and  also  the  square  of  the  last,  b"^.  Thus,  the 
second  term  being  found,  the  repetition  of  a  similar  process  will 
evidently  lead  to  the  root,  whatever  number  of  terms  the  ex- 
pression may  contain.     Hence  the  following 

Rule.  Arrange  the  terms  according  to  the  powers  of  some 
letter,  and  extract  the  root  of  the  first  term,  which  must  be  a 
cube,  or  some  power  of  a  cube  ;  place  this  root  in  the  quotient, 
subtract  its  cube  from  the  first  term,  and  there  will  be  no  re- 
mainder. 

Bring  down  the  three  next  terms  for  a  dividend,  and  put 
three  times  the  square  of  the  root  just  found  in  the  divisor^s 
place,  and  see  how  often  this  is  contained  in  the  first  term  of  the 
dividend,  and  the  quotient  is  the  next  term  of  the  root. 

Add  three  times  the  product  of  the  two  terins  of  the  root,  plus 
the  square  of  the  last  term,  to  the  term  already  in  the  divisar^s 
place,  and  the  divisor  will  be  completed. 

Multiply  the  complete  divisor  by  the  last  term  of  the  root ; 
subtract  the  product  from  the  dividend,  and  to  the  remainder 
connect  the  three  next  terms,  and  proceed  as  before. 

EXAMPLES. 

1.  Find  the  cube  root  o?  a^-]-Za%^Sab''-{-b\ 

a'-\-U%-{-Zab^-\-b\a-^b. 


U''+Ub-\-b^)-\-Sa'b-\-Sab''-\-b 


2.  Extract  the  cube  root  o^ x^—'Sx^-{-bx^—Sx—l. 

x^—Sx^J^bx^—Sx—l{x^—x—l. 


—82-^+82:^—  x^ 


82:^_62:^4-32:-f-l)-82:^+62-3— 82:-! 
— 32:^-1-62:5-32:— 1. 


12 


134  ALGEBRA. 

The  first  divisor  is  found  thus  : 

And  the  second  thus  : 

S[x'—xY-\-3{x'—x)  {-l)-^{—lY=Sx'—Qx^-\-Sx-{-l. 

3.   Extract   the   cube  root  of  x^—Qx^-\-lbx'—20x^-\-lbx^— 
62'4-l. 

x^—Qz'^ldx'—20x'-\-lbx''-Qx-{-l{x''—2x-\-l, 
x' 


Sx'—Qx^+4:x'')—Qx'-\-lbx'—20x 
-Qx'-^12x'—  8x 


Sx'—12x^-\-lbx''—Qx+l)Sx'—12z'-\-'^bx''—QxJrl 

Bx'-123^+lbx''—6x-{-l. 


4.  Extract  the  cube  root  of  x^-\-dx'^-\-27x-{-27, 

5.  Extract  the  cube  root  of  1—6 2/+ 12?/^— 8?/^ 

6.  Extract  the  cube  root  of  «^— 6a^+40a^— 96a— 64. 

Ans.  a^—2a—4:. 

7.  Extract  the  cube  root  of  a^-{-Sa'b-{-Bab''-{-b^-^Sa'c-\-Qabc 
-\-Sb''c-{-Bac'^-i-Sbc^-{-c\  Am.  a-f^+c. 

BY   DETACHED    COEFFICIENTS. 

1.  What  is  the  cube  root  of  x'-\-Qx'—^^x^-\-^^x—U  ? 
l_|_6+0-404-0+96-64(l+2-4    ■ 


rx3  =        3)     6 


(1+2)3=  i_|_6_[-l2+  8 

1^x3  =  a)_i2_48 


1+6-1-  0-40+0+96—64, 
Hence,  1+2— 4=.t'^+22:— 4.     Ans. 


EVOLUTION.  135 

2.  What  is  the  cube  root  o^  Sx^—BQx' -{-^4x^—27x^1 
8-]-0-364-0+54-|-0-27(2-[-0-3. 


2-X3  =    12)4-0—36 


84-0-36+0-1-54 -f-0— 27. 
Hence,  24-0— 3=2z^4-0ar^— 32:=2:c3— 3z.' 

3.  What  is  the  fourth  root  of  a;*4-8a;^-f-24a;24-32:r4-16  ? 

Am.  x-\-2. 

4.  What  is  the  cube  root  of  x^—2>x^y-\-'^x^'f — ?/^? 

Ans.  x^ — y. 

187.  Reasoning  analogous  to  that  employed  in  Art.  185  will 
show,  that,  if  a  point  be  placed  over  every  third  figure,  begin- 
ning at  the  unifs  place,  the  number  of  points  thus  placed  will 
be  the  number  of  digits  in  the  cube  root ;  and  attention  to 
Art.  186  will  furnish  the  following  operation : 

1.  Extract  the  cube  root  of  1860867. 

.     a  -\-  b  -\-c 
1860867(1004-204-3=123. 
a?  =  1000000  =  first  subtrahend. 


So"  =       30000)860867  =  first  remainder. 


da'b  =  600000 

^ab''  =       120000 

b'  =  8000 


728000  =  second  subtrahend. 


3(^4-3)2  =  43200)132867  =  second  remainder. 

S{a+bYc  =  129600 

3(^4- 3)c2  =  3240 

c^=  27 


132867  =  third  subtrahend. 

This  process  is  the  origin  of  the  Rule  given  on  page  248  of 
the  Author's  National  Arithmetic,  to  which  the  pupil  is  re- 
ferred. 


136  ALGEBRA 


SYMBOLICAL  FORM. 

1860867(100+204-3 
(100)^=    1000000    [=123. 


3(100)2+3(100)2+(20)^=36400)860867 

728000 


3(100+20)2+3(100+20)3  +  3^=44289)132867 

132867. 


2.  What  is  the  cube  root  of  31255875  ?  A7is.  315. 

3.  What  is  the  cube  root  of  37259704  ?  Ans.  334. 

4.  What  is  the  cube  root  of  116930169  ?  A7is.  489. 

5.  What  is  the  cube  root  of  508.169592  ?  Am.  7.98. 

6.  What  is  the  cube  root  of  .724150792  ?  Am.  .898. 

188,  To  extract  any  root  of  a  compound  algebraical  quantity. 

Since  {a-}-x)'"=a"'-]-7na'^~^x-\-  &c.,  it  is  obvious,  that  when 
the  quantities  are  properly  arranged,  and  the  first  term  of  the 
root  is  found,  the  second  term  of  the  ?nih.  root  will  be  obtained 
by  dividing  the  second  term  of  the  proposed  quantity  by  ?na""~\ 
or  by  m  times  the  first  terra,  raised  to  the  (vi — l)th  power. 

And,  if  the  root  thus  found  be  raised  to  the  Tnih.  power,  and 
the  result  be  subtracted  from  the  quantity  proposed,  and  the 
process  be  repeated  when  necessary,  any  root  of  a  compound 
quantity  may  be  determined. 

The  similarity  of  the  processes  employed  in  this  and  the  pre- 
ceding articles  will  be  immediately  noticed,  it  being  observed  in 
the  former,  the  complete  powers  of  a  monomial,  binomial,  tri- 
nomial, &c.,  are  subtracted  from  the  proposed  quantity  by  07ie, 
two^  three^  &c.,  operations;  whereas,  in  the  latter ^  the  subtrac- 
tion of  the  same  quantities  is  efi'ected  at  07ice.  Hence  the 
following 

General  Rule.  1.  Arrange  the  terrm  so  that  the  highest 
power  shall  stand  in  the  first  term,,  and  let  the  7iext  higher  occupy 
the  second  place. 

2.  Firid  the  root  of  the  first  ter7n,  and  place  it  in  the  quotient ; 


EVOLUTION.  137 

and,  having  raised  this  root  to  the  required  power,  subtract  it 
from  the  first  term,  and  then  bring  down  the  second  tern  for  a 
dividend. 

3.  Involve  the  root  last  found  to  the  next  inferior  power,  and 
multiply  it  by  the  index  of  the  given  power  for  a  divisor. 

4.  Divide  the  dividend  by  the  divisor,  and  the  quotient  will  be 
the  next  term  of  the  root. 

5.  Involve  the  whole  root  thus  found  to  the  required  power, 
which  subtract  from  the  given  quantity,  and  divide  the  first  term 
of  the  remainder  by  the  same  divisor  as  before. 

6.  Proceed  in  this  manner  for  the  next  term  of  the  root,  and 
so  proceed  until  the  work  is  finished. 

See  page  255  of  the  Author's  National  Arithmetic. 

EXAMPLES.     . 

1.  Required  the  square  root  of  a*— 2a^a;-f3aV — ^a^-^-x"^, 
a'—2a'x-{-Sa''x^—2ax^-\-x\a''—ax+x\ 


2a')—2a'x 


a'—^a^x^a^x"- 


2a2)2aV 


a'—la^x-^Za^x'—^ax^-^-x'. 

2.  Required  the  cube  root  of  a;^-j-6:c^— 40:c^+96a:— 64. 

ajS-j-e^:^— 40:?;^+962:— 64(:c^+2:c— 4. 
x^ 


%x')M' 


a;6_|.6:c^_[_12:^44_8^3 


3:c^)— 12.^^ 


a:6-|_6a;^— 40^:34-962:— 64. 


3.  Required  the  fourth  root  of  \^x''—^^xSj^2VoxY—'21^^'if 
^^\y\ 

12# 


138  ALGEBRA. 

Wx'—mx'y+21Qxy—21Qxf+Sly\2x—dy. 
16x' 


S2x')—mx'y 


lQx'-9Qx'y-lr21Qxy—21Qxf+Sh/. 

4.  Required  the  cube  root  of  w^ — Q?n^-{-4cOm^ — 96m — 64. 

Atis.  iri} — 27a — 4. 

5.  Required  the  fifth  root  of  322;'— 8 0^:^+8 0:^:^—4 0x^+1 0:e 
— 1.  Ans.  2x — 1. 


SECTION   XV. 

SURDS,    OR   RADICAL   QUANTITIES. 

Art.  189t  Surds,  or  radical  quantities,  are  roots  whose  values 
cannot  be  exactly  obtained,  being  usually  expressed  by  means  of 
the  radical  sign,  or  fractional  indices ;  in  which  latter  case  the 
numerator  shows  the  power  to  which  the  quantity  is  to  be  raised, 
and  the  denominator  its  root. 

I  2 

Thus,  a/S^  or  3^,  denotes  the  square  root  of  3.     s/d\  or  a^, 

m 

is  the  cube  root  of  the  square  of  a;  and  c",  or  s/oJ^t  is  the  wth 
root  of  the  ?7zth  power  of  a. 

190.  The  quantity  /y/^,  or  a/3",  is  an  irrational  quantity  or 
surd,  because  no  number,  either  whole  or  fractional,  can  be 
found,  which,  when  multiplied  by  itself,  will  produce  either  2  or 
3  ;  but  their  proximate  values  may  be  found,  to  any  degree  of 
exactness,  by  the  common  rule  for  extracting  the  square  root. 

Problem  I. 

191.  To  reduce  a  rational  quantity  to  the  form  of  a  surd,  or 
radical  quantity. 


RADICAL     QUANTITIES  139 

KuLE.  Raise  the  quantity  to  a  power  corresponding  to  the 
index  of  the  surd  to  which  it  is  to  be  reduced^  and  over  this  neio 
quantity  place  the  radical  sig7i,  or  proper  index,  and  it  ivill  be 
the  form  required. 

EXAMPLES. 

1.  Let  5  be  reduced  to  the  form  of  a  square  root. 
Here  5x5=5^=25;  whence  V^-     -^^^« 

2.  Reduce  2x'^  to  the  form  of  the  cube  root. 

Here  {2xY=Sx' ;  whence  ,^^/^,  or  {Sx'fy  or  S^x'^.  Ans. 

3.  Let  — 2x  be  reduced  to  the  form  of  the  cube  root. 
Here       {—2xY=  —  Sx^;  therefore  4^— 8^1     Ans. 

4.  Let  Sa^  be  reduced  to  the  form  of  the  square  root. 


x^ 
5.  Let  —  be  reduced  to  the  form  of  the  cube  root. 


Am.       1^. 


6.  Reduce  x^  to  the  form  of  the  fifth  root.  An^. 

x- 
x—y 


x" 
7.  Let  be  reduced  to  the  form  of  the  fourth  root. 


Am.    (-^\^. 
\{x-yf) 

8.  Let  (x — y"')  be  reduced  to  the  form  of  the  square  root. 

Alls.  ({x—y'^f\^. 

192.  If  a  rational  quantity  be  joined  to  a  surd,  it  may  be 
reduced  to  the  form  of  a  surd  by  raising  the  rational  part  to  the 
required  power,  and  multiplying  it  by  the  surd. 

9.  Let  5 a/ 7  be  reduced  to  a  simple  radical  form. 

5^^=V5x5X^^=A/^X^/7=>^/^75■.    Ans. 

10.  Let  ^Asfa  be  reduced  to  a  simple  radical  form. 

3Va=V3x3xV«=/>/^     ^ns. 


140  ALGEBRA. 

11.  Let  3/^3  be  reduced  to  a  simple  radical  form. 
3A^=/^3x3x3X/v^=/V^XAf3=/s^/^.     Aiis. 

12.  Let  ^/s/a  be  reduced  to  a  simple  radical  form.  A?is.  a/J. 

13.  Let  4/^yF  be  reduced  to  a  simple  radical  form. 


4 

Atis. 


16' 


14.  Let  Z/ym  be  reduced  to  a  simple  radical  form. 

Ans.  /</3"7?i. 

15.  Let  =-     — — -  be  reduced  to  a  simple  radical  form. 


x-\-l    \x—l_    [p  +  iy/a:— 1\_    \x'-\-x'—x—l 

.T— 1-^^+1"  ^V^—iy  \x-]-i)~  >^x^—x''—x-\-i' 


— i— ,  or  ( -y.     Ans, 

X—l  \X—1J 


16.  Let  —    U— 2  be  reduced  to  a  simple  radical  form. 

A...   Jt' 

Problem  II. 

193.  To  reduce  quantities  of  difierent  indices  to  others  that 
shall  have  a  given  index. 

Rule.  Divide  the  indices  of  the  quantities  given  by  the  index 
under  which  the  quantities  are  to  be  reduced,  and  the  quotients 
will  be  the  new  indices  for  those  quantities. 

Then,  over  the  quantities  with  their  new  indices  place  the 
given  index,  and  they  will  be  the  equivalent  quantities  required. 

EXAMPLES. 

1.  Reduce  4^  and  8"^  to  other  quantities  of  the  same  value, 
each  having  the  common  index  ^. 


RADICAL   QUANTITIES.  141 

Here  J-i-i=^Xf=l=3»  t^®  fii'st  index. 

And-  ^-j-^=^Xf=3=^'  ^^®  second  index. 

Whence        (4^)^=4^ ;  and  (8^)^=8^     Atis, 

194.  The  truth  of  this  rule  will  be  evident;  for  if  4  be  raised 
to  the  3d  power,  and  the  6th  root  extracted,  that  root  will 
be  equal  to  the  square  root  of  4. 

Thus,         4x4x4=64;  ..^/154=2  ;  /v/4=2. 

And,  if  8  be  raised  to  the  2d  power,  and  the  6th  root  extracted, 
the  result  will  be  equal  to  the  cube  root  of  8. 

Thus,  8x8=64;  /y64=2;  ^8=2. 

2.  Reduce  3^  and  5^  to  the  common  index  ^. 

Ans.  a^2^a79;  ^=,^/1M. 

3.  Reduce  a^  and  <z^  to  quantities  that  shall  have  the  common 
index  ■^.  Ans.  A/l^  and  ts/~^' 

4.  Reduce  3a^  and  2a*  to  the  quantit^s  that  shall  have  the 
common  index  ^.  Ans.  3^a^  and  24/a^. 

1  J. 

5.  Reduce  hx^  and  62/^   to  quantities  having   the   common 

index  yV-  ^ns.  h)^'!?  and  6]^^. 

m  p 

6.  Reduce  a"  and  b^  to  quantities  having  a  common  index  ~. 

H  JL  Z  1. 

Therefore  a"=(a'"^)"^;  and  b^={b"P)"^. 

Problem  III. 

195.  To  reduce  surds  to  a  common  index. 

Rule.  Reduce  the  indices  of  the  quantities  to  a  common  de- 
nominator^ and  then  involve  each  quantity  to  the  power  denoted 
by  its  numeratoi'. 


142  ALGEBRA. 

EXAMPLES. 

1.  Reduce  3^'  and  4^^^  to  quantities  having  a  common  index. 

We  first  reduce  the  fractional  indices,  |-  and  ■^,  to  a  common 
denominator,  and  find  them  to  be  §  and  |,  which  have  the  same 
value  as  J-  and  -^. 

Hence  3^=3^=  (32)6=27^  or  ^^27. 

And  4^=4^=(4~)'^=:16"^  or  />/T6. 

2.  Reduce  4^  and  6*"  to  equal  quantities,  that  shall  have  the 
same  index. 

1  and  ^  =  3-^2-  and  ^. 

Therefore  4:^==:4^^={A')^'^={2b6)'^'^  or  V2F6.     Arts. 
And  6^=6^^=(6^)i^=(216)'^'2-  ^r  jy^TO:     Ans. 

2  _1 

3.  Reduce  2^  and  3-  to  equal  quantities  having  a  common 
iJidex.  jins.   /^OB'and /^yST. 

-i  -1 

4.  Reduce  a^  and  b^  to  equal  quantities  having  a  common 

Mex.  ^^.    /y^and/^/^ 

5.  Reduce  x'"  and  ?/"  to  quantities  having  a  common  index. 

Ans.  '""V^  and  "^"A/y^. 

Problem  IV. 

196.  To  reduce  surds  to  their  most  simple  form. 

Rule.  Resolve  the  given  quantity  into  two  factors,  ojie  of 
which  shall  be  the  greatest  corresponding  power  contained  in  it, 
and  set  the  root  of  this  power  before  the  remaining  factoi',  with 
the  proper  radical  sign  between  them. 

Note. — When  the  given  surd  contains  no  factor  which  is  an  exact 
power,  it  is  already  in  its  most  simple  form.  Thus  a/ 15  cannot  be  re- 
duced lower,  because  neither  of  the  factors  5  or  3  is  a  square. 

examples. 
1.  Let  a/IB  be  reduced  to  its  most  simple  form. 


RADICAL     QUANTITIES.  143 

We  divide  48  into  two  factors,  16  and  3,  16  being  the  great- 
est power  of  the  required  root.  We  therefore  extract  the 
square  root  of  16,  and  write  its  root,  4,  before  the  other  factor, 
having  the  sign  prefixed  to  the  surd. 

Thus  V48=a/T6X3==4V3.     Ans. 

2.  Let  /v^TOB  be  reduced  to  its  most  simple  form. 

In  this  question  we  find  the  factors  of  108  to  be  27  and  4, 
27  being  the  largest  possible  factor  of  which  the  cube  root  could 
be  extracted.     The  operation,  therefore,  is 

Thus  a7108=4/27X4=3/^:     A?is. 

3.  Let  a/To  be  reduced  to  its  most  simple  form. 

Am.  5v^- 

4.  Let  /v^SO  be  reduced  to  its  most  simple  form. 

A71S.  2^. 

5.  Reduce  /s/TTc^  to  its  simplest  form. 

Here  V  27aV= V  9^^^*  X  ^clx= //9a^  X  /s/^ax= ^ax^A/^ax. 

6.  Reduce  /^fh^^cc'or}  to  its  simplest  form.  Ans.  ^ax/^2a'^x. 

Problem  V. 

197.  When  any  number  or  quantity  is  prefixed  to  the  surd, 
that  quantity  must  be  multiplied  by  the  root  of  the  factor,  as  in 
Art.  196,  and  the  product  must  then  be  joined  to  the  other  part, 
as  before. 

EXAMPLES. 

1.  Let  2/\A32  be  reduced  to  its  most  simple  form. 

Here  2^32=2^/16x2=2 x4/v/2=8/v/2:    A7is. 

In  performing  this  question  we  first  find  the  factors  of  32, 
which  are  16  and  2. 

We  then  extract  the  square  root  of  16,  and  multiply  its  root, 
4,  by  the  number  prefixed  to  the  surd,  and  find  the  product  to 
be  8,  to  which  we  subjoin  the  surd  2. 

19S.  This  and  all  similar  questions  might  have  been  per- 
formed by  squaring  the  number  prefixed  to  the  surd,  and  then 


144  ALGEBRA.  * 

multiplying  this  number  by  the  surd.  Let  this  product  be 
divided  into  two  factors,  as  before,  and  the  square  of  the  former 
prefixed  to  the  latter  will  give  the  answer. 

Thus,  2/v/32=V2X^X32=V128=a/134x2=:8a/2.  Ans, 

2.  Let  b^^/^4:  be  reduced  to  its  most  simple  form. 

Here  5^24=5A/8x3=5x2/^^=10A^'3: 


Or   5a^24=a^/^X5x5><24=/>^3000=^1000x3=10.^/^ 
3.  Reduce  2/^/W  to  simple  terms.  Ans.  4/v^ 

Problem  VI. 

199.  A  fractional  surd  may  be  reduced  to  a  more  convenient 
form  by  multiplying  both  the  numerator  and  denominator  by 
such  a  number  or  quantity  as  will  make  the  denominator  a  com- 
plete power  of  the  kind  required,  and  then  proceeding  as  be- 
fore. [Art.  198.] 

EXAMPLES. 

1.  Let  a/^  be  reduced  to  its  most  simple  form. 

2.  Let  /^/~%  be  reduced  to  its  most  simple  form. 

3.  Let  /\/y  be  reduced  to  its  most  simple  form. 

Ans.  |a/T4. 

4.  Let  /^/J  be  reduced  to  its  most  simple  form. 

Ans.  i/V^. 

5.  Let  />/J  be  reduced  to  its  most  simple  form. 

Ans.  |/v^. 

EXAMPLES   TO   EXERCISE   THE   FOREGOING    RULES. 

1 .  What  is  the  most  simple  form  of  a/1%  ?       An^.  b/s/W. 

2.  What  is  the  most  simple  form  of  V  80aV  ? 

Ans    \axsPSx. 


RADICAL    QUANTITIES.  145 

3.  What  is  the  most  simple  form  of  /^/T^^aWc^l 

Atis.   2>ab\/  lac^. 

4.  What  is  the  most  simple  form  of  7/\ASD  ?     Ans.  28^/57 

5.  What  is  the  most  simple  form  of  |A/f  ?       Ans.  -^^a/^. 

6.  What  is  the  most  simple  form  of  xxa/^  ?     Ans.  -^-j/^/T. 

7.  Let  /\/96a-r^  be  reduced  to  its  most  simple  form. 

Ans.  4iax/s/Wx, 


8.  Let  ^/v^^SB^F+Glp  be  reduced  to  its  most  simple  form. 

Ans.  ^/^(lx'-\-^f). 

Problem  VII. 

200 1  To  add  surd  quantities  together. 

I.  When  the  radicals  are  similar,  annex  the  radical  part  to 
the  sum  of  the  coefficients. 

EXAMPLES. 

1.  Add  7/v/2  to  5  V2:  Ans.  12^/57 

2.  Add  b/^~ah  to  Z/s/~ab.  Ans.  S/s/ab. 

3.  Add  a/s/xy  to  hfy/xy.  Ans.  {a-\-b)/s/lcy. 


4.  Add  "J^d^—y  to  y/s/d'—y.  Ans.  (7+?/)/y/a^ZI^. 

II.  When  the  radical  parts  are  dissimilar,  make  them  similar 
by  Art.  197,  and  proceed  as  above. 

But,  if  the  surd  part  cannot  be  made  the  same  in  all  the 
quantities,  they  can  only  be  added  by  the  signs  +  and  — . 

5.  Add  a/T5  and  a/Z2  together. 

First  VT8=V  9x2=3/v^. 

And  /v/^=,v^l6x~2=4v^. 

Then  3V^+4V2=7>\/2:     Ans. 

6.  Required  the  sum  of  />^375  and  /v^I92. 
First  a^^^7B=a^125x3=5a^. 
And  /^/Tm=^  64x3=4^/3: 
Then  54/T-|-4.^y3=9.^.     Ayis. 

13 


146  ALGEBRA. 

7.  Required  the  sum  of  a/27  and  /\/4H.  ^^-  7V^. 

8.  Required  tlie  sum  of  />/S(J  and  .a^T^.        Am.  11 V^- 

9.  Find  the  sum  of  V^I^  and  VioF.  A7is.  15a/3T 

10.  It  is  required  to  find  the  sum  of  /^/40  and  y^l35. 

il;i5.  5/v^. 

11.  Find  the  sum  of  4^/^  and  54/128.         ^?i5.  324/2. 

12.  Find  the  sum  of  ^/fand  ^/S-  ^^-  S^ 

13.  Required  the  sum  of  %s/ d^h  and  5a/16^. 

Problem  VIII. 
201 1  To  find  the  difiierence  of  surd  quantities. 

Rule.  When  the  radicals  are,  or  have  been  made,  similar, 
aiinex  the  common  radical  part  to  the  difference  of  the  rational 
parts. 

But,  if  the  quantities  have  no  commo7i  surd,  they  can  be  sub- 
tracted only  by  changing  the  sign  of  the  subtrahend. 

examples. 

1.  From  VMTtake  ^./W. 

First  //320=a/(I4x^=8V5T 

And  ^/~8D=A/I(Jx5=4^/5: 

Then  8V^-4^5'=4a/5:    Ans. 

2.  Find  the  difference  between  /v/T28"  and  a/^. 
First  A7I2B=4/(5ix2=44/2. 
And  4/~54=4/27x2=34/5: 
Then  44/^-34/2=a/2:     Ans. 

3.  Required  the  difference  between  2.a/^  and  /\/TK. 

Am.  7a/2T 

4.  What  is  the  difference  between  24^213  and  3a/3^  ? 

^7W.    24/5. 


RADICAL    QUANTITIES.  147 

5.  llequired  the  difference  of  VTo  and  a/?S^     A7is.  a/3] 

6.  Required  the  difference  of  /v/256  and  /^82. 

Am.  2a/4: 

7.  Required  the  difference  of  a^  and  a/^ . 

Am.    ^/s/~^. 

8.  Required  the  difference  of  /^^  and  a^/^^ 

Ans.  -j2^/^/T5. 

9.  Find  the  difference  of  f  A^o^and  f  ,</^^ 

10.  From  ts/  ^.ax^  take  3.?;A/I)a.  ^?w.  — 7a: a/^ 

Problem  IX. 

202 1  To  multiply  surd  quantities  together. 

Rule.  When  the  surds  are  of  the  same  kind,  find  the  -product 
of  the  rational  parts,  and  the  product  of  the  surds  ;  and  the  two 
joined  together,  with  the  common  radical  sign  between  them,  will 
give  the  ivhole  product  required,  which  may  be  reduced  to  its 
most  simple  form  by  Art.  199. 

203.  If  the  surds  are  of  different  kinds,  they  must  be  reduced 
to  a  common  index,  and  then  multiplied  together,  as  before. 

204.  Powers  and  roots  of  the  same  quantity  are  multiplied 
by  adding  their  exponents. 

EXAMPLES. 

1.  Find  the  product  of  3/s/B  and  2a/6. 
Here  3a/B 
Multiplied  by              2a/6 

Gives  6A/iK=6A/(16x3)=24/sA3:  Am. 

2.  Find  the  product  of  ^^/f  and  f  4/|7 


148  ALGEBRA. 

Here  ^a^ 

Multiplied  by  f /^ 

Gives  |^=  f  a/  (#Xf )  =  W  m)  == 

3.  Multiply  2^  by  3*. 

Here  2^= 2'6 =(2^)^=8^'. 

And  3^=3^=(32)^=9^ 

72^.     ^715. 

4.  Multiply  5  v^  by  3/^ 

1  3 

Here  bA/a=ba^=ba^, 


i 


And  3^=3a  =3a^ 


5 


lba^=15/^.     Am. 

5.  Multiply  4VT2  by  3a/2.  ^tz^.  24^^. 

6.  Multiply  3/v/2  by  2v^.  ^tz^.  24. 

7.  Multiply  1^  by  f^^/TF.  tItz^.  ^^/B". 
V  8.  Multiply  tv^  by  i^^a/I:  ^;z5.  fVlV 

9.  Multiply  7.^0:3  by  S^/IT  ^?Z5.  70^^. 

10.  Multiply  i^  by  A^^J^T.  ^tw.  ^V/^^^T02- 

11.  Multiply  2a^  by  a^.  A7is.  2a\ 

12.  Multiply  («+3)^  by  (a-j-b)^.  Ans.  '1/ {a^Vf\ 

13.  Multiply  x—s/xy-\-y  by  j\fx-\-s/y. 

By  expressing  the  surds  with  fractional  indices,  we  have 

3  XX 

x^ — xy^-\-x^y 
-^xy'^—x-y-^-y^ 

3  3 


RADICAL    QUANTITIES.  149 

14.  Multiply  aP-^-a'b^+Jb^-^-ab-i-a^b^+b^  by  a^—b^. 
J + a"b^-{-ah^^ab~\-Jb^-\-  b^ 
a^—b^ 


a^^ah^-i-a'b^-\-ah^ab^-JraV^ 

5     1  2  3  4  15 

—an'^—a'b'^—an—ab^—an^—b^ 
a^  ^.     Am. 

15.  Multiply  /s/a-{- a/I + a/c  by  sfa-^/sfb — sfc. 

Ans.  a-{-b — c-\-2/>y~ab. 

Problem  X. 

205  •  To  divide  one  surd  quantity  by  another. 

E,uLE.  When  the  surds  are  of  the  same  ki?id,  JiTid  the  quotient 
of  the  ratioTial  parts^  and  the  quotients  of  the  surds,  and  tfie  two 
joined  together^  with  the  common  radical  sign  between  them,  will 
give  the  whole  quotient  required. 

But,  if  the  surds  are  of  different  kinds,  they  must  be  reduced 
to  a  common  index,  and  be  divided  as  above. 

The  quotients  of  different  powers  or  roots  of  the  same  quantity 
are  found  by  subtracting  their  indices. 

EXAMPLES. 

1.  Divide  Q/s/m  by  3V^ 

Here?^^=2^A2=2^/4X^=(2x2)A/B=4V3:    A71S. 
Sa/S 

2.  Divide  ^^/IM  by  2a/^. 

Here?^^^=4^Arg=4V^X2=(4x3)V2=12V^.  Ans. 
2a/1 

3.  Divide  8^"512  by  4^. 


Here  ?^^^=2/>^/^5^=2/y64x4=8A^     Ans, 
4a/2 

13# 


150  ALGEBRA. 

4.  Divide  12  times  the  cube  root  of  280  by  3  times  the  cube 
root  of  5. 

Here  ^^"^^^^ =4:,yM=4:y^/^yO = SaTT.     Am. 
3/^/5  ^ 

5.  Divide  6a/51  by  3a/2.  Am.  6V^. 

6.  Divide  4;.^/T2  by  2^0^:  Ans.  2,^. 

7.  Divide  4/,/Mhj  2^/W.  Ans.  2VW: 

8.  Divide  64/100  by  3,^/5'.  Ans.  2a^'^. 

9.  Divide  /v/20+VT2  by  a/5+V^  ^^5.  2. 

10.  Divide  32fAA'by  IBfA^  Ans.  ^^[-Y, 

206,  Since  the  division  of  surds  is  performed  by  subtract- 
ing their  indices,  it  is  evident  that  the  denominator  of  any 
fraction  may  be  taken  into  the  numerator,  or  the  numeratoi 
into   the   denominator,  by   changing   the  sign  of  its  index. 

EXAMPLES. 

1.  Let  -  be  expressed  by  a  negative  index. 

a      1 

2.  Let  —  be  expressed  by  a  negative  index. 

3.  Let  —  be  expressed  by  a  negative  index. 

4.  Let  —  be  expressed  by  a  negative  index. 

^2—    1    — ^      • 


RADICAL     QUANTITIES.  151 

5.  Let  a  ^  he  expressed  by  a  positive  index. 

_ 1      a~^      1 

6.  Let      be  expressed  by  a  negative  index. 

Ans.  {a-{-x)~2. 

7.  Let  a{a^ — x^)~'^  be  expressed  by  a  positive  index. 

1 

Ans.  — ^ -i. 

8.  What  is  the  value  of  — ? 

or 

or- 

Whence  it  follows  that  a"  is  a  symbol  equivalent  to  unity ; 
consequently  1  may  always  be  substituted  for  it.  This,  how- 
ever, has  been  demonstrated  in  a  previous  article. 

Problem  XI. 

207 •  To  involve  or  raise  surd  quantities  to  any  power. 

h 
Let   a°    represent  a   surd  quantity;    then,  by  Art.  204,  its 
square  will  be 

Therefore,  to  involve  a  surd  to  any  required  power,  we  adopt 
the  following 

KuLE.  When  the  surd  is  a  simple  quantity,  multiply  its 
index  hy  'ifor  the  square,  3  for  the  cube,  SfC,  and  it  will  give 
the  power  of  the  surd  part,  which,  being  annexed  to  the  proper 
power  of  the  rational  parts,  ivill  give  the  whole  power  required. 

If  the  surd  be  a  compound  quantity,  multiply  it  by  itself  the 
requisite  number  of  times. 

EXAMPLES. 

1.  What  is  the  square  of  3a^  ? 

^a^^^=U^=^^a\     Ans, 


152  ALGEBRA. 

2.  What  is  the  cube  of  f  a/3  ? 

Here  (|V3)'=/y//57=^VV(9X3)=|^/^.     Ans. 

3.  Required  the  square  of  3/^3.  Ans.  9/^/Vi. 

4.  Required  the  cube  of  17/v^^         Aiis.  103173^21. 

5.  What  is  the  fourth  power  of  ^/v/B"?  Am.  -jJ^. 

6.  Required  the  cube  of  y\/F'  Am.  Sa^/W. 

7.  Required  the  third  power  of  ^V^  Ans.  ^/s/^. 

8.  Required  the  fourth  power  of  ^/«/2T  A7is.  ^. 

1  "* 

9.  What  is  the  mth.  power  of  a"  ?  Am.  a". 

10.  Required  the  square  of  2-{-a/W.  Am.  7+4/v/^ 

r  I.  ^ 

11.  What  is  the  -th  power  of  a''  ?  Am.  a?\ 

Problem  XII. 

208.  To  find  the  roots  of  surd  quantities. 

Rule.  When  the  surd  is  a  simple  quantity,  multiply  its 
index  by  ^  for  the  square  root,  hy  ^for  the  cube  root,  c^c,  and  it 
will  give  the  root  for  the  surd  part,  which  being  annexed  to  the 
root  of  the  rational  part,  will  give  the  whole  root  required. 

The  truth  of  this  rule  may  be  illustrated  by  the  following 

EXAMPLES. 

1.  What  is  the  cube  root  of  the  square  root  of  64  ? 

The  square  root  of  64= VM=6#=8. 

And  the  cube  root  of  8=a/B=8^=:2.     Ans. 

209.  The  same  result  would  have  been  obtained  if  we  had 
multiplied  the  index  (^)  of  the  given  quantity  by  the  index  of 
the  required  root  (^),  the  product  of  which  is  4X-^=i  5  and  if 
we  had  considered  this  (-^)  the  index  of  the  root  to  be  extracted 
of  the  given  quantity  64,  the  operation  would  have  been  thus : 

/y^=2.     Ans.,  as  before. 


RADICAL     QUANTITIES.  153 


2.  Required  the  cube  root  of  the  square  root  of  a. 

^2  /\  ^=a^.     Atis, 

3.  Required  the  fourth  root  of  /v/BT        3^^^=3^.     Am, 

4.  What  is  the  square  root  of  O/vAo"? 

Here  (9^3)2=9^X3*'^^=9^X3^=34/^ 

5.  What  is  the  square  root  of  10^  ? 

10^=1000  ;  VT000=10V1U.     Ans. 

6.  What  is  the  cube  root  of  ||^/\/^?  Atis.  ^/^/~a. 

7.  What  is  the  square  root  of  ^f  a^  ?  Atis.  %c^/sj~a. 

Probleji  XIII. 

210.  To  find  factors  that  shall  cause  any  surds  to  become 
rational. 

I.  When  the  surd  is  a  monomial,  multiply  it  by  the  same 
quantity,  with  an  index  such  as  when  added  to  the  index  of  the 
given  quantity  will  make  it  a  unit. 

The  quantity  A/a~or  a-  is  made  rational  by  multiplying  it  by 
\J a  or  a-. 

Thus,  f/a^s/a^  or  arY^a^z=a. 

-1  2. 

And  it  will  be  rational  if  a^    be   multiplied   by   a^,    thus, 

i  2 

4  \ 

Also,  if  a^    be  multiplied  by  a^  it  will  be  rational;  thus, 
k       1 

EXAMPLES. 

2  1 

1.  What  factor  will  make  x'^  rational  ?  Ans.     a;"^. 

2  5 

2.  What  factor  will  make  y^  rational  ?  ^7i5.  y^ . 

3.  What  factor  will  cause  ar^  to  become  rational  ? 


154  ALGEBRA. 

II.  When  the  surd  is  a  binomial  or  residual  quantity,  and 
both  the  terms  are  even  roots,  to  find  a  factor  that  will  make  the 
quantity  rational. 

In  Art.  158  we  have  shown  that  the  product  of  the  sum  and 
difference  of  any  two  quantities  is  equal  to  the  difference  of 
their  squares ;  therefore,  when  one  or  both  of  the  terms  are 
even  roots,  we  multiply  the  given  binomial  or  residual  by  the 
same  quantity,  with  the  sign  of  one  of  its  terms  changed. 

Note.  —  It  is  sometimes  necessary  to  repeat  the  operation. 

EXAMPLES. 

1.  To  find  a  multiplier  or  factor  that  shall  make  4-{-/vA5 
rational. 

Griven  surd,  4-{-/,/5 

Multiplier,  4— />/5 


16-f-4^/^ 


Product,  16  — 5=11  rational  quantity. 

2.  Find  a  factor  that  shall  make  /s/a-\-/sfh  rational. 

^ — b 


a  — b  rational  quantity. 

3.  What  factor  will  make  l-|-/vA5~i*^tional  ? 

1-V^ 
l+VB 

— V  o — 6 


—  8= — 2  rational  quantity. 


R  A  U  I  C  A  L     Q  U  A  N  T  I  TI  K  S .  155 

4.  What  factor  will  make  a/o  — a/1  rational  ? 

/V^-HVI 


5— a/o 
-f-V^-1 

5  — 1=4  rational  quantity. 

5.  Find  multipliers  that  shall  make  A/b-\-/^/^  rational. 

/>/B+a/3 
a/o-^"B 

a/5  -a/B 

5-Vlo 
+A/I5-3 

5  — 3=2  rational  quantity. 

6.  What  multiplier  will  make  /\/o  —  />/rFrational  ? 

a/o— a/^ 
a/O+a/S 

5 — a/o^ 
+  a/ox — a; 

5  — X  rational  quantity. 

III.  A  trinomial  surd  may  be  rendered  rational  by  changing 
the  sign  of  one  of  its  terms  for  the  multiplier. 

EXAMPLES. 

1.   To   find   multipliers    that    shall    make    /\/7+a/B— a/^ 
rational. 


156  ALGEBRA. 


7+V21-A/14 
+/vA21+3-a/(3 

8+2a/2I 
-8+2a/21 


-64-16a/21 

+16^/2I+84 
84 — 64=20  rational  quantity. 

2.  Find  a  factor  that  will  make  a/^— a/1— V^  rational. 
a/^-a/1-VB 

8~a/B-a/21 

+//24-/s/a-3 

4-2V3 

4+2V^ 

16-8a/3 

+8/v^— 12 
16—12=4  rational  quantity. 

QUESTIONS   FOR   EXERCISE. 

1.  Find  a  multiplier  tliat  shall  make  a/^  — V^  rational. 

Atis.  V5+/v/^. 

2.  Find  a  multiplier  that  shall  make  a/7+a/(j  rational. 

Ans.  /v/7— V^- 

3.  Find  a  multiplier  that  shall  make  .^/ 10— a/2  rational. 

A77S.  a/TU+a/2'. 


RADICAL     QUANTITIES.  157 

4.  Find  multipliers  that  shall  make  /s/a-\-/s/h-\-A/c  rational. 

Am.  s/a—h/h—f>/^,  and  {a—h—c-^-^sTbc). 

5.  Find  multipliers  that  shall  make />y^—/v/lL  rational. 

Am.  (/4^+^1)(a/^+/\A). 

Problem  XIV. 

Art.  211.  To  reduce  a  fraction,  whose  denominator  is  a 
surd,  to  another  that  shall  have  a  rational  denominator,  without 
changing  its  value. 

Rule  1.  When  the  proposed  fraction  is  a  simple  one,  multiply 
each  of  its  terms  by  the  denominator. 

2.  If  it  he  a  compound  surd^  find  such  a  multiplier  hy  the  last 
Art.  as  will  make  the  denominator  rational,  then  multiply  both 
the  numerator  and  denominator  by  it. 

EXAMPLES. 

1.  Reduce  to   a   fraction   whose   denominator   shall   be 

/>/a 

b    ^,/\/~a     J/s/fl        . 
rational.  X— — = -•     Am. 

fs/a     /sfa        ^ 

2  Reduce  to   a  fraction   whose   denominator   shall  be 

rational.  -^X— ~=^^^-     ^^^s- 

^a     f^        « 

2 

3.  Reduce  the  fraction  —    to  another  whose  denominator 

Vo 

shall  be  rational.  2         2   _/v/5     2/v/5       . 

= — ^X — r=— ^.     Am. 

/\/5     a/5     V^5        ^ 

3 

4.  Reduce  to  a  fraction  whose  denominator  shall 

/s/^-a/2 
be  rational. 

^^3                   3  VH-a/^    3a/F+3^/2 

Here  = — —z X —  ^  — 


sj^sj2     j^-fsf2     a/o+/V^  ^-^ 

3VF+3V2-^VF-fV2-^^^^^     ^^. 
3  1 

14 


158  ALGEBRA. 

5.  Extract  the  square  root  of  |. 

Here     F^^^^^^^^^^_^^^,    ^,,. 

6.  Reduce  r 7=  to  a  fraction  whose  denominator  shall  be 

rational. 

Here     V^   ^    V^      3+V^     3V2+2_3V2+2_ 


3-^     3-V^     3+V^        9-2  7 

7.  Reduce  — — to  a  fraction  that  shall  have  a  rational 

denominator.  ^^w.   ^ . 

8.  Reduce  7 7^  to  an   equivalent  fraction  having  a  ra- 

3 — a/1 

tional  denominator.  Ans 


2    * 

5 

9.  Reduce  the  fraction to  an  equivalent  fraction 

a/5-V^ 
having  a  rational  denominator.  Ans.   ^ . 

10.  Reduce  the  fraction to  an  equivalent  fraction 

having  a  rational  denominator.  Ans.   = . 

1 

11.  Reduce      ^  . — —  to  a  fraction  that  shall  have  a  ra- 

V^+a/7 

.      ,  ,         .                                                     A        a/^— a/7 
tional  denominator.  JiJis.    — x- — • 


# 


RADICAL     QUANTITIES.  159 

3 

12.  Reduce  the  fraction to  an  equivalent  fraction 

that  shall  have  a  rational  denominator. 

Am.   J5 =V^+V ^• 

o 

Problem  XV. 

212*  To  change  a  binomial,  or  residual  surd,  into  a  general 
surd. 

lluLE.  Involve  the  given  hiTWinial,  or  residual,  to  a  power 
corresponding  with  that  denoted  hy  the  surd;  then  write  the 
radical  sig7i  of  the  same  root  over  it. 

EXAMPLES. 

1.  It  is  required  to  reduce  24-/>/S  to  a  general  surd. 
Here,  (2-{-VF)^=4+4V"5+B=74-4V^. 
Therefore,  2+VE=V(7+4V3). 

2.  Reduce  /v^+a/^  to  a  general  surd. 

Here,  (V2+V3f=2+2v^4-3=5  +  2VI>. 

Therefore,  a/^+V3=a/(o-|-2V^- 

3.  Reduce  ^/2-\-/^/~i  to  a  general  surd. 

Here,  (/^+..yi)^=6+64/^-j-6^. 

Therefore,  a/2+a^=aJ^+a^+>^). 

4.  Let  3 — a/5  be  reduced  to  a  general  surd. 

Ans.  //(U-GVo). 

5.  Let  /v/2'+ 2/\/^  be  changed  to  a  general  surd. 

Ans.  /s/|2(3+SV^3). 

6.  It  is  required  to  change  4 — vT  to  a  general  surd. 

Ans.  //(23— 8VT). 

7.  Let  1  /sf^—o^s/^  be  changed  to  a  general  surd. 

Ajis.  /^^/(786-13234^-f567^¥). 


160  ALGEBRA. 

Problem  XVI. 

TO   EXTRACT   THE   SQUARE   ROOT   OF    A   BINOMIAL   SURD. 

213.  A  binomial  surd  is  one   in  which  one  of  the  terms,  at 
least,  is  irrational ;  as  a-\-\/F,  or  fJ~a-\- k/T. 

To  extract  the  square  root  of  (z-fV^s  we  put 
/s/  {a-\-  j\fh)=^Tn-\-n. 
And  fs/  {a—^/V)^=m—n. 

By  squaring  both  of  these  equations. 

We  have  a-\-fs/T=irr^-\-%nn-\-7^. 

And  a — /s/b=7)r — 2mn-\-7i?. 

By  addition,        la         =2^^         -^2n^. 
And  a=z7rt^-{-7i^. 

Multiplying  the  two  first  equations  together, 

We  have      //(«+VT)X  V(fi^— //^)  =  (^+^)X(w— ^). 

And  //(^^ — h)=m^ — tz^ 

Having  both  the  sum  and  difference  of  m}  and  n^,  we  obtain, 
by  addition  and  subtraction,  the  following  equations  : 

>^-=°+^f-^-l,  and  „.^'^-Vf-^). 
Therefore,  ^=v(2±^f^"^), 

Consequently,  V{a+V*)=V(^i^^— )  + 
/a-V(a'-^)\ 

And  V(a-V^)=v(^-±^^-^')-V(^-=^-=^). 

It  is  certain  that  both  a  and  /^{a^—l)  must  be  rational,  in 
order   that   the    expressions   within   the    parentheses   may  bo 


RADICAL     QUANTITIES.  161 

rational,  in  which  case  each  of  the  above  values  will  be  either 
two  surds,  or  a  rational  and  a  surd. 

The  above  formula)  will  apply  to  any  particular  values  for  a 
and  h ;  observing  that  if  h  be  negative,  the  signs  of  h  in  the 
formula)  must  be  changed. 

EXAMPLES. 

1.  What  is  the  square  root  of  11-|-a/T2  ? 
Here,  a=ll,  and  i=72.     Therefore, 

And     v(°-^^--l))=v("-^f '-^^'  )=V2. 

Therefore,  V(ll4-/v/7^)=34-V^ 

2.  What  is  the  square  root  of  10— a/M? 
Let  a=10,  and  3=96. 

Then  v("-±^?^)=v(i^^°^^)=^ 

And    v(-^-^))==v(^"-^f-^^')=2. 

Therefore,  ^(\S)-s/M)=s/^—% 

3.  ^Vhat  is  the  square  root  of  6-f-/y/2D?  Ans.  l-|-/\/5. 

4.  What  is  the  square  root  of  6-f-2V~5?  Ans.  s/h-\-\. 

5.  What  is  the  square  root  of  12-|-2^"3o  ?  Ans.  V^-fVi'- 

6.  Required,  the  square  root  of  36±10^Tr7~ 

Ans.  SrhVlT. 

7.  What  is  the  square  root  of  7— 2VT0  ?     Ans.  ^o— /s/2. 

8.  What  is  the  square  root  of  1+4^"^^? 

Ans.  2-1-^/^=:^,  or  2-,v^:=^. 

14^^ 


162  ALGEBRA. 

SECTION    XVI. 

IMAGINARY    QUANTITIES. 

Art.  214.  As  every  algebraical  symbol  hitherto  considered, 
whether  it  be  affected  with  the  sign  -}-  or  — ,  when  raised  to 
an  even  power  gives  a  positive  result,  it  follows  that  no  even 
root  of  a  negative  quantity  can  be  either  positive  or  negative. 
The  even  roots  of  negative  quantities  having,  therefore,  no  sym- 
bolical representation  in  accordance  with  the  views  of  Algebra, 
so  far  as  we  have  yet  considered  it,  can  only  be  indicated  or 
expressed  by  means  of  the  radical  sign,  or  corresponding 
fractional  inder.  Hence  arises  a  new  sjjecies  of  symbolical 
expressions,  called  Imaginary  or  Impossible  Quantities. 

Thus  the  square  root  of  — a^  is  neither  -\-a  nor  — a,  but  is 
written  V  —d\  and  is  equivalent  to  />/<^^X( — l)=/>/^V — 1 
=dz«V'— Ij  which  is  said  to  be  impossible,  or  imaginary,  in 
consequence  of  involving  the  symbol  a/  —  1. 

By  Art.  78  we  learn  that  the  product  of  real  quantities,  that 
have  like  signs,  is  always  plus  ;  and,  if  the  signs  are  unlike,  the 
product  is  minus.  We,  therefore,  infer,  that  the  product  of  two 
imaginary  quantities,  that  have  the  same  sign,  is  equal  to 
minus  the  square  root  of  their  product,  considering  them  as  real 
quantities. 


Hence,         (+a/— «)(+/%/— ^)= — sj  cL--=i—a. 


(—a/ — a)(— V — a)  =  — /\/(2-=— a. 

(— /\/— a)( — js/ — h)  =  —/s/ab. 

215.  If  the  two  imaginary  quantities  have  different  signs, 
then,  it  is  evident,  their  product  will  be  equal  to  plus  the  square 
root  of  their  product,  considering  them  as  real. 

Thus,         (-|-V^Z^)(— V^=^)  =  -|-V^. 


IMAGINARY     QUANTITIES.  163 

EXAMPLES. 

1.  Multiply  4^^=B  by  IsT^. 


2.  Multiply  ^-^rsT—^  by  3— V— o- 

3— //"=F 


12+3V^^ 

8.  Multiply  3V^^  by  7a/^^.  ^^-  — 21v^. 

4.  Multiply  — 7V^=¥by  -3.^^=^.        ^tw.  -21^/T2: 

'  5.  Multiply  44-a/^=^  by  V^^-      ^^^5-  4V'=^-/\^T5'. 

216.  If  one  imaginary  be  divided  by  another,  having  the 
same  signs,  the  quotient  is  equal  to  plus  the  square  root. 

But,  if  the  imaginaries  have  different  signs,  it  is  evident  that 
their  quotient  will  be  equal  to  minus  the  square  root  of  their 
quotient. 

EXAMPLES. 

6.  Divide  6V^=S  by  2v'=i:  Am.  3Vf . 

7.  Divide  Isf^-^  by  — 5/v^^^.  Ans.  — |/>/o. 

1 


8.  Divide  —/sf—l  by  — 7v^^^.  Ans.    +^^7tt- 

9.  Divide  -\-,sJ  —a  by  -\-s/  —h.  Ans.    ■\-/sJj' 

1) 


a 


10.  Divide  —s/  —a  by  —s/—h.  Ans.    -\-\/-j. 


h 


11.  Divide  44- V"^^  by  2— a/^=^.  Ans.  1+//^"^- 

12.  Divide  1-f  V^l  by  1— V^=^I.  Ans.  sf^^. 

13.  Divide  2V^^  by  — 3^^=^-  ^^^-  — f V|^ 


164  ALGEBRA. 

SECTION    XVII. 

QUADKATIC  EQUATIONS,  OR  EQUATIONS  OF  THE  SECOND 

DEGREE. 

Art.  217.  A  quadratic  equation  is  one  in  whicli  the  un- 
known quantity  rises  to  the  second  power. 

Quadratics  are  of  two  kinds:  those  which  contain  only  the 
square  of  the  unknown  quantity  are  called  pure  quadratics,  and 
those  which  contain  both  the  first  and  second  powers  of  the 
unknown  quantity  are  called  affected  quadratic  equations. 

The  following  are  examples  of  pure  quadratics : 

EXAMPLES. 

1.  Given  4:x'—7=29  to  find  x. 
Conditions,  42;^— 7=29. 
Transposing,                    4a;2=29+7=36. 
Dividing,                          x'^=9. 
Extracting  square  root,  x=-\-^. 

2.  Given  ax'^-\-b=c  to  find  x. 

Conditions,  ax'^-{-b=c. 

Transposing,  ax^=c — b. 

Dividing,  x^=zc — b. 


Extracting  square  root,  a;=-4- 


a 
\c—b 


4 


a 

Hence,  to  find  the  value  of  the  unknown  term,  we  have  the 
following 

Rule.  Transpose  and  reduce  the  equation,  so  that  the  un- 
known quantity  may  be  positive,  and  the  first  member  of  the 
equation.  Divide  both  members  of  the  equation  by  the  coefficieiit 
of  the  unknown  quantity ;  then  extract  the  square  root  of 
both  members. 


QUADRATIC  EQUATIONS.  165 

3.  Given  5a:'+5=3a;-+55  to  find  x. 
Conditions,  5:^;^-f-5=3a;^-{-55. 
Transposing,                   5z- — 3a;'^=55— 5. 
Reducing,                       22;-= 50. 
Dividing,  x'^=2b. 
Extracting  square  root,   x —  1  5. 

4.  Given  2a;2+ 8=3.^2—28  to  find  x. 
Conditions,  3.i"— 28=2a;2+8. 
Transposing,  Sz^— 2a;'=28  +  8. 
Reducing,  x^=oQ. 
Extracting  square  root,  a:  =  -l-6. 

5.  Given  7a;2— 5=32:^+11  to  find  x.  Am.  a;=±2. 

6.  Given  4x'-\-lb=7x''—4:17  to  find  x.        Ans.  x=±:\2. 

hx~ 

7.  Given  32-^+7=^+35  to  find  x.  Ans.  2:= ±4. 

8.  Given  ax^-\-n^m — c  to  find  x.   Ans.  a;=  + 


9.  Given  x- — ab=.d  to  find  x.  Ans.  x=-h\/d-\-ab. 

10.  A  lady  bought  a  silk  dress  for  £8  15^.,  and  tlie  number 
of  sbillings  she  paid  per  yard  was,  to  the  number  of  yards,  as 
4  to  7.  How  many  yards  did  she  purchase  for  her  dress,  and 
what  was  the  price  per  yard  ? 

Let  X  =  the  number  of  shillings  paid  per  yard. 

Ix 
Then  -J-  =  the  number  of  yards. 

7^2 
And  the  price  of  the  whole,  ~-r-=  175  shillings. 

Clearing  of  fractions,  7a;-=700. 

Dividing,  a;'^=100. 

Extracting  the  square  root,  x=.\^s.,  price  per  yd. 

Ix 
Therefore,  —=17^  yards.     A7is. 


166  ALGEBRA. 

11.  I  have  10  acres  of  land.  If  it  were  a  square  field,  what 
would  be  the  length  of  one  of  its  sides  ?  Ans.  40  rods. 

12.  A  and  B  lay  out  money  on  speculation  ;  the  amount  of 
A's  stock  and  gain  is  $27,  and  he  gains  as  much  per  cent,  on 
his  stock  as  B  lays  out.  B's  gain  is  $32 ;  and  it  appears  that 
A  gains  twice  as  much  per  cent,  as  B.  Required  the  capital  of 
each.  Am.  A's  capital,  $15;  B's,  $80. 

13.  There  are  two  square  fields,  the  larger  of  which  contains 
25,600  square  rods  more  than  the  other,  and  the  ratio  of  their 
sides  is  as  5  to  3.     Required  the  contents  of  each. 

Ans.  Contents  of  the  larger,  40,000  square  rods. 
Contents  of  the  smaller,  14,400  square  rods. 

14.  I  have  three  square  house-lots,  of  equal  size ;  if  I  were  to 
add  193  square  rods  to  their  contents,  they  would  be  equivalent 
to  a  square  lot  whose  sides  would  measure  each  25  rods.  Re- 
quired the  length  of  each  of  the  sides  of  my  three  house-lots. 

Ans.  12  rods  each. 

15.  A  farmer  has  a  square  field,  and  the  number  of  rods 
round  it  is  -^h  the  number  of  square  rods  of  its  contents.  Re- 
quired the  number  of  acres  in  the  field.  A7is.  10  acres. 

16.  John  Smith  has  a  field,  which  is  a  right-angled  parallel- 
ogram ;  its  sides  are  in  the  ratio  of  4  to  3 ;  a  diagonal,  passing 
from  one  corner  to  its  opposite,  is  100  rods.  Required  the 
contents  of  the  field.  Ans.  30  acres. 

17.  Two  workmen,  A  and  B,  engage  to  work  for  a  certain 
number  of  days,  at  different  rates.  At  the  end  of  the  time,  A, 
who  had  been  absent  4  days,  received  75  shillings ;  but  B,  who 
had  been  absent  7  days,  received  only  48  shillings.  Now,  if 
B  had  been  absent  only  4  days,  and  A  7  days,  they  would  have 
received  exactly  alike.  How  many  days  were  they  engaged  for, 
how  many  did  each  work,  and  what  had  each  per  day  ? 

Atis.  They  were  engaged  to  work  19  days.  A  worked  15, 
and  B  12  days ;  A  received  5  shillings,  and  B  4  shillings  per 
day. 


QUADRATIC     EQUATIONS.  1(37 

18.  Two  numbers  are  to  each  other  as  4  to  5,  and  the  sum  of 
their  cubes  is  1512.     What  arc  those  numbers  ? 

Ans.  8  and  10. 

19.  A  bushel  measure  contains   2150f  cubic  inches,  and  I 
wish  to  make  a  box  that  shall  contain  50  bushels.     Its  lengt] 
is  to  be  to  its  breadth  as  3  to  1,  and  its  height  f  its  breadth. 
What  are  its  dimensions  ? 

A71S.  Length  108.84-f ,  breadth  36.28+,  and  height  27.21  + 
inches. 

20.  What  must  be  the  dimensions  of  a  cubical  box  that  shall 
contain  100  bushels  ? 

A?is.  Height,  length,  and  breadth,  59.9+  inches. 

21.  Two  numbers  are  to  each  other  as  3  to  7,  and  the  differ- 
ence of  their  cubes  is  2528.     What  are  those  numbers  ? 

Atis.  6  and  14. 

22.  Bought  a  house-lot  for  S5184.  Its  length  is  to  its  breadth 
as  3  to  1.  I  gave  as  many  dollars  per  square  rod  as  the  lot  is 
rods  in  breadth.     What  were  the  dimensions  of  the  lot  ? 

yins.  36  rods  long,  12  rods  wide. 

Problems. 

23.  Let  771  be  divided  into  two  parts,  whose  squares  shall  be 
to  each  other  as  7i  to  p. 

Let  X  =  the  greater. 

And  m — x  =  the  less. 

Then  x"^  :   (wi — xY  :  :  n  :  p. 

Multiplying  extremes,       px^=.7i{7n — xY. 

Evolution,  x,s/p=dz/^yn{j}i—x). 

Reducing,  xs/p^=m\fn—x,,/n. 

Transposing,      X\/~p-\-Xs/Ti^mspri. 

-p..  .,.  ms/n 

Dividmg,  2-= the  greater. 

\fpArs/n 

Subtracting,    m = the  less. 

sfp-^sj^i     \fp-\-fs/n 


168  ALGEBRA. 

If  we  take  the  minus  sign,  we  have 

x/>/p=.  — s/n{m — X). 
Multiplying,  x^=—m/s/n-\-x/s/~n. 

Transposing,        Xfy/p—x/s/n=  —  m^7i 
Changing  signs,  X/^i—x/sf^^^m,s/n. 


771 /</n.      . 
Dividing,  x=^-—^ — —  the  greater, 

'    ,          ,                           ms/n           —msfp 
Subtracting,  m = the  less. 

24.  Divide  18  into  two  such  parts  that  the  square  of  the 
larger  part  shall  be  25  times  the  square  of  the  less. 
Let  X  =  the  larger;  then  18 — x  =  the  less. 
Then  we  have  x^  :   (18-2f  :  :  25  :  1. 

Multiplying  extremes,        a;^=25(18— :r)^. 
Evolution,  a; =5  (18— a-). 

Multiplying,  x=.%^—hx. 

Transposition,  6a;=90. 

Dividing,  :r=15,  the  larger. 

18—15=3,  the  less. 

VERIFICATION. 

152=25(3)1 
Involving,  225=225. 

THE   THEORY   OF    THE   LIGHTS    AND   ATTRACTION. 

218.  To  apply  the  foregoing  problems,  we  premise  the  fol- 
lowing principles  of  Natural  Philosophy. 

1.  The  intensity  of  light  emanating  from  any  luminous  body 
is  inversely  as  the  square  of  the  distance  from  that  body  ;  that 
is,  if  the  earth  were  twice  the  distance  from  the  sun  that  it  now 
is,  it  would  receive  only  one-fourth  part  of  the  light  and  heat 
that  it  now  does  ;  and,  if  it  were  removed  to  ten  times  the 
distance,  it  would  have  only  one-hundredth  part  of  the  light  and 
heat. 


QUADRATIC     EQUATIONS.  169 

2.  The  quantity  of  light  emanatiDg  from  a  celestial  body  is 
directly  as  the  square  of  its  diameter. 

Hence,  if  the  earth  were  four  times  the  diameter  of  the  moon, 
an  inhabitant  of  that  luminary  would  receive  sixteen  times  as 
much  light  from  the  earth  as  he  would  receive  from  the  moon 
if  he  were  on  the  earth. 

3.  The  laws  of  attraction  are  similar  to  those  of  light,  for  all 
bodies  attract  each  other  inversely  as  the  squares  of  the  dis- 
tances from  their  centre,  and  directly  as  the  masses  of  matter 
which  compose  those  bodies. 

APPLICATION   OF    THE   ABOVE   PRINCIPLES. 

25.  The  moon  is  240,000  miles  from  the  earth,  and  the 
quantity  of  matter  in  the  earth  is  80  times  that  of  the  moon. 
At  what  distance  from  the  earth,  in  a  direct  line  towards  the 
moon,  must  a  body  be  placed  to  be  equally  attracted  by  each,  so 
that  it  will  remain  at  rest  as  it  respects  those  bodies  ? 

Let  d  =  the  distance  between  the  moon  and  earth. 
e  =  the  quantity  of  matter  in  the  earth. 
7?i  =  the  quantity  in  the  moon. 

X  =  the  distance  from  the  earth  to  the  point  required. 
Then  d — x  =  the  distance  from  the  moon. 
We  have  then  the  following  proposition  : 

As  x"^  :  {d — xY  :  :  e  :  m. 

Therefore,  mx^=e((l — xY. 

By  evolution,  x/>/m-=fsJT{d — x). 

Reducing,  x/^  m=^ds/~e^xs/~e. 

Transposing,  x/s/mA^xs/  e=.d\/~e. 

JDividmg,  x-=.- 


Substituting  the  value  of  d^  e  and  ?/^,  we  have 
240,000  V  BO      2146624.8 


V80+V1       8.94427+1 
tance  from  the  earth. 
15 


=215865.4  miles,   =  the   dis- 


170  ALGEBRA. 

240,000— 215865.4=24134.6  miles,  =  the  distance  from  the 
moon.  / 

If  we  take  the  negative  sign,  we  shall  find  the  point  beyond 
the  moon  where  the  attraction  of  the  two  bodies  will  be  equal. 

Taking  the  minus  sign,         xs/m= — /s/~e{d — x). 
Reducing,  x\/m==i — dfsfe-\-x/s/~e. 

Transposing,  xjsfe — x/s/Tn=.d\/~e~. 

Uiviamg,  x-=. —  . 

Substituting  the  values  of  d,  e  and  m,  we  have 

2146624  8 

x=    .^-^ ^—=270,210  miles  from  the  earth's  centre,  and, 

a/oO — V  1 

therefore,  270,210—240,000=30,210  miles  beyond  the  moon. 

26.  Required  the  distance  from  the  earth,  in  a  direction  tow- 
ards the  sun,  where  a  body  would  remain  at  rest,  the  distance 
of  the  earth  being  95,000,000  miles  from  the  sun,  and  the 
quantity  of  matter  in  the  sun  being  333,928  times  greater  than 
that  of  the  earth. 

Let  S  represent  the  quantity  of  matter  in  the  sun,  E  the 
quantity  of  matter  in  the  earth,  and  D  the  distance  between  the 
earth  and  sun,  and  x  the  required  distance  from  the  sun. 

Then,  substituting  these  letters  for  those  in  question  23,  we 
have  the  following  formula  : 

d/\/J~ 

95,000,000V3M;928^^^^33^^33^. 

V  ^33,928+  vr 

95,000,000—94,835,885=164,115  miles.    Am. 

27.  The  diameter  of  Venus  is  7700  miles,  its  distance  from 
the  sun  is  68,000,000;  the  diameter  of  the  earth  is  7912  miles, 
and  its  distance  from  the  sun,  as  stated  above,  is  95,000,000 
miles.     How  much  greater,  therefore,  is  the  intensity  of  light  at 


QUADRATIC  EQUATIONS.  171 

Venus  than  at  the  earth,  and  what  is  the  comparative  quantity 
that  each  receives  from  the  sun  ? 

A71S.  The  intensity  of  light  at  Venus  is  1.95-|-  tunes  greater 
than  at  the  earth.  Venus  receives  from  the  sun  1.84-f-  times 
more  light  than  the  earth. 

28.  Mercury  is  37,000,000  miles  from  the  sun.  How  much 
greater,  therefore,  is  the  intensity  of  light  and  heat  at  Mercury 
than  at  the  earth  ?  Ans.  ^-f-^^^  times. 

29.  Jupiter  is  490,000,000  miles  from  the  sun,  and  its  diam- 
eter is  89,000  miles.  Saturn  is  900,000,000  miles  from  the 
sun,  and  its  diameter  is  79,000  miles.  How  much  more  light, 
therefore,  do  we  receive  from  Jupiter  than  from  Saturn,  when 
they  are  in  opposition  to  the  sun  ? 

Let  a  =  the  distance  of  Jupiter  from  the  sun. 

b  =  the  diameter  of  Jupiter, 
c  =  his  distance  from  the  earth. 
d  =  the  distance  of  Saturn  from  the  sun. 
e  =  the  diameter  of  Saturn. 
h  =  his  distance  from  the  earth. 

The  distance  of  these  planets  from  the  earth  is  obtained  by 
subtracting  the  earth's  distance  from  the  sun  from  their  dis- 
tance from  the  sun. 

The  surface  of  Jupiter  is  to  the  surface  of  Saturn  as  the 
squares  of  their  diameters ;  and  as  the  quantity  of  light  which  a 
planet  receives  from  the  sun  is  as  the  square  of  its  diameter  di- 
rectly, and  inversely  as  the  squares  of  its  distance  from  the  sun, 

Therefore,  if  b'^  =  the  surface  of  Jupiter, 
and         e^  =  the  surface  of  Saturn, 
and         a  and  d  their  respective  distances  from  the  sun, 
then  the  intensity  of  light  at  Saturn  will  be  to  the  intensity 

of  light  at  Jupiter  as  -,,  is  to  — .     And  as  the  light  which  each 

of  these  planets  gives  to  the  earth  is  in  intensity  inversely  as 
the  squares  of  their  distances  from  the  earth, 

therefore,  if—  =  the  quantity  of  light  at  Saturn,  and    —  = 
d"  a^ 


172  ALGEBRA. 

n 


the  quantity  of  light  at  Jupiter,  then  -r—^  ==  the  quantity  of 


light  which  Saturn  gives  to  the  earth,  and  -y^  =  the  quantity 

which  Jupiter  gives. 

Therefore,  to  find   how  much   more   light   we   receive   from 
Jupiter  than  from  Saturn,  we  use  the  following  proportion : 

Therefore,  x=  .,  „  „. 

If  we  substitute  for  these  letters  their  numerical  values,  we 
shall  have 

900^X805^X89^ 


X=i. 


79'^X490'X395' 
810000X648025X7921 


=17.7+.     Am. 


6241X240100X156025 

That  is,  we  receive  more  than  seventeen  times  as  much  light 
from  Jupiter  as  we  do  from  Saturn. 

In  the  above  operation,  we  have  cancelled  the  ciphers  in  the 
distances  and  diameters  of  the  planets, 

AFFECTED   QUADRATIC    EQUATIONS. 

219.  An  afiected  quadratic  equation  is  one  containing  the 
first  power  of  the  unknown  quantity  in  one  term,  and  the  square 
of  that  quantity  in  another  term. 

Every  equation  of  this  kind,  having  any  real  or  positive  root, 
will  fall,  when  properly  reduced,  under  one  of  the  four  following 
forms : 

2.  x^—ax^z     b,  where  2=-}-^riiV  (  jH"^  )• 

3.  x'^-{-ax=z—b,  where  2-=  — -±V(t — ^)« 

,   d  ,  f  CL'  \ 

4.  x-—ax=—h,  where  x=-{---±:/\/  \t~^)' 


QUADRATIC  EQUATIONS.  173 

220t  No  exact  root  can  be  taken  of  a  binomial ;  but,  if  the 
first  term  of  a  binomial  be  a  square  of  the  unknown  quantity, 
and  the  second  term  the  quantity  itself,  with  1,  or  any  other 
quantity,  for  its  coefficient,  the  square  of  half  the  coefficient  of 
the  second  term,  added  to  the  binomial,  will  make  the  whole 
quantity  an  exact  square.  This  may  be  illustrated  by  the  fol- 
lowing examples. 

Let  x^-\-4:X  be  the  binomial,  then  2  is  half  the  coefficient  of 
the  second  term,  and  its  square  is  2x2=4.  This  we  add  to 
the  binomial,  and  the  result  is  :c--f  4a;-[-4,  and  this  quantity  is 
an  exact  square,  and  its  root,  by  Art.  183,  is  x~\-2. 

If  the  binomial  be  x^-\-ax,  and  we  add  to  it  the  square  of 

half  the  coefficient  of  x,  — ,  the  sum  will   be  x^-\-ax-}~j-,  the 

exact  root  of  which  is  x-{--. 

Again,  if  the  binomial  be  x^ — Zahx,  we  have  only  to  add  the 

Q„2I2 

square  of  half  the  coefficient  of  a;,  which  is  — — ,   to   the    bino- 

Q    2/2 

mial,  and  the  sum  will  be  an  exact  square,  x^ — Zahx-\ . 

For  (  x-—oabx-\ — j-  y  =x — . 

221.  If,  therefore,  there  be  any  binomial  whose  first  term  is 
an  even  power  of  the  unknown  quantity,  and  the  second  term 
half  that  power,  and  we  add  the  square  of  half  the  coefficient  of 
the  second  term  to  the  binomial,  the  result  will  be  an  exact 
square. 

222.  To  solve  an  affiicted  quadratic  equation,  we  adopt  the 
followino' 

O 

Rule,  Bring  all  the  unknown  terms  to  one  side  of  the  equa- 
tion, and  the  known  terms  to  the  other,  observing  so  to  arrange 
them  that  the  term  which  contains  the  square  of  the  unknown 
quantity  shall  be  positive,  and  stand  first  in  the  equation,  and  the 
term  luhich  contains  the  first  power  of  the  unknown  quantity  the 
second  term  of  the  equation. 
15=^ 


174  ALGEBRA. 

Divide  each  side  of  the  equation  by  the  coefficient  of  the  un- 
known square. 

Add  the  square  of  half  the  coefficient  of  the  second  term  to  each 
side  of  the  equation^  and  the  unknown  side  imll  he  a  complete 
square. 

Extract  the  square  root  of  each  side  of  the  equation,  and  from 
the  result  the  value  of  the  unknown  quantity  may  he  obtained. 

Given  x^-\-^x=^^i  to  find  the  values  of  a:. 

Here,  by  the  question,  7? -\-%x-=.%\. 

Completing  the  squares,  2;^-}-8:r-|-16=84-f-16=100. 

Extracting  the  square  root,       a:-f-4=10. 

Whence,  rr=10— 4. 

And,  x=.\i.     Ans. 

In  solving  this  question,  we  first  add  the  square  of  half  of  8, 
that  is,  16,  to  both  sides  of  the  equation ;  we  then  extract  the 
square  root  of  x^-j-^^+l^'  ^^<^  ^^^  ^^  result  to  be  a:+4,  and 
the  square  root  of  100=10.  Therefore,  a;-j-4=10,  that  is, 
a:=10— 4=6.     Ans. 

223,  It  may  also  be  demonstrated,  by  the  following  diagram, 
that  if  the  square  of  half  the  coefficient  of  the  second  term  be 
added  to  the  first  member  of  an  equation,  it  will  be  a  complete 
square. 

Let  x  represent  one  side  of  the 
square  ABCD ;  then  x^  will  represent 
this  square.  To  this  square  we  must 
add  82:,  and  this  quantity  must  be  ap- 
plied equally  to  the  two  sides  AB 
and  BC,  or  the  figure  would  not  be 
a  square.  Therefore  4x,  which  is  half 
of  82:,  will  be  applied  to  either  side,    d 


If  this  quantity,  4a:,  be  divided  by  x,  the  quotient,  4,  will  repre- 
sent either  of  the  distances  EA  or  BGr.  Having  added  the  two 
equal  parallelograms  EABF  and  BGHC  to  the  square  ABCD, 
we  find  our  figure  needs  the  small  square  FBGL  to  complete  the 
square.     The  contents  of  this  must  be  equal  to  the  product  of 


QUADRATIC  EQUATIONS.  175 

FB  and  BG,  that  is,  4  multiplied  by  4,  or  the  square  of  4  = 
16 ;  but  4  is  half  the  coefficient  of  the  second  term.  We  add 
this  quantity  to  x^-^Sx,  and  the  sum  is  ar^-{-8:c-j-16,  and  its 
square  root  is  x-\-4:,  by  Art.  182. 

224.  A  quadratic  may  be  solved  by  the  following 

Rule.  Having  transposed  the  unknown  terms  to  one  side  of 
the  equation,  and  the  known  to  the  other,  multiply  each  side  by 
4  times  the  coefficient  of  the  square  of  the  unknown  quantity. 

Add  the  square  of  the  coefficient  of  the  first  power  of  the  un- 
known quantity  to  both  sides  of  the  equation,  and  the  unknown 
side  will  then  be  a  complete  square. 

Extract  the  root  of  both  members,  and  the  value  of  the  un- 
known quantity  is  obtained  as  before. 

EXAMPLES. 

1.  Given  Sa^'^+^a:— 7=88  to  find  the  values  of  a:. 
Conditions,  ^x^-\-^x—l^'^'^. 
Transposing,  3a:2+4a:=88+7=95. 
Multiplying  by  4  times  3,  36f2+48a:=1140. 
Completing  the  square,   36a;24-48a;+16=1140+16=1156. 
Evolving,  6:?:-l-4=±34. 
Transposing,  62'=±34— 4=30,  or— 38. 
Dividing,  x=b,  or  —6^. 

2.  Given  2a;'^— 10a;-f  7=— 5  to  find  the  values  of  a:. 
Conditions,  2^^— 10a;4-7=— 5. 
Transposing,  2x^ — 10a;=— 5 — 7  =—12. 
Multiplying  by  4  times  2, 16a;-— 80:c=— 96. 
Completing  the  square,    16a;^-80a:+ 100=— 96+100=4. 
Evolving,  4a;— 10=±2. 
Transposing,                                   4a;=±2+10=12,  or  8. 
Dividing,  a;=3,  or  2. 

3.  Given  3a:--f-5a;— 8  =  34  to  find  the  values  of  a;. 

Arts.  xz=zo,  or  — 4|. 


176  ALGEBRA. 

4.  Given  z'^-{-Qx-{-4:=22 — x  to  find  the  values  of  a:. 

Atis.  x=2y  or  — 9. 

5.  Given  Sx'^ — 72:4-6  =  171  to  find  the  values  of  x. 

Atis.  x=5,  or  — ^^-. 

Yj^x 350 

6.  Given j-lOz — 20=175  to  find  the  values  of  x. 

X 

Ans.  x=7,  or  — 5. 

7.  Given  :c~ — 6a; -{-12=4  to  find  the  values  of  x. 

Atis.  x=4c,  or  2. 

8.  Given  Sx'^-i-S2x=2Q0  to  find  the  values  of  x. 
Conditions,  Sx''-{-S2x=SQ0. 
Dividing,  a;'^-}~4:c=45. 
Completing  the  square,    a;--j-4:i;-|-4=45-{-4=49. 
Evolving,  2:-}-2=±7. 
Transposing,  x=zt7 — 2=5,  or  — 9. 

9.  Given  a;"— 8a;-{-50=:98  to  find  the  values  ofx. 

Conditions,  2:''— 8.^+50=98. 

Transposing,  a;'— 82:=98— 50=48. 

Completing  the  square,  x"^ — 82:-}-16=48-|-16=64. 
Evolving,  2;— 4 =±8. 

Transposing,  a:=±8-f-4=12,  or  —4. 

10.  Given  x'-\-ax=zb  to  find  the  values  of  a;. 
Conditions,  x'^-\-ax=b. 

Completing  the  square,  x'^-}-ax-\—T-=b-{'-r-. 


a 


Evolving,  ^-{-■^=± 


Transposing,  a:=i: 


('4)- 


11.  Given  ox"^ — 3a;-j-6=5^  to  find  the  values  of  a:. 
Conditions,  3a;'— 3a:-)-6=5^. 


QUADRATIC     EQUATIONS.  177 

Transposing,  ^z^ — 3a:=5^— 6= — f. 

Reducing,  x^ — x= — f. 

Completing  the  square,    x^—x-\-l= — |-|-i=-}-^Jg-. 
Evolving,  x—^=dzi' 

Transposing,  2-=±^-|-2-=§'  o^'  h 

2 

12.  Given  | — ^+20^=42f  to  find  the  values  of  x. 
Conditions,  — f_|_20i=42f. 

Transposing,  ^-|=42| -20^=22^. 

Clearing  of  fractions,     x^ — ^=44^. 

2x1  1     400 

Completing  the  square,  a:- — -}--==44^-J-^=-Tr— . 

1         20 


Evolving,  x—-=^—=±^. 

o  o 

Transposing,  a;=zt:6|4-^='<''5  oi'  —  6|. 

13.  Given  c2:'^-|-Z'J!;=c  to  find  the  value  of  x. 
Conditions,  ax^-\-bx=^c. 

Dividing,  x'^A =-. 

^        ,    .        ,                     ^     bx       b'^       c       b'- 
Completing  the  square,  x'^-\ [--— ^= — b-r-^- 

€L        "iCti        Qj       ^Cb 

\/c        P\        b 
Evolving  and  transposmg,  a:=±      (  — f-j^  J  —  — 

14.  Given  ax" — bx-{-c=d  to  find  the  values  of  x. 
Conditions,  ax'^—bx-{-c=d. 
Transposing,  ax'^ — bx=d — c, 

-p..  .,.                                  2     ^^     ^— ^ 
Dividing,  x' = . 


za 


178  ALG1]BRA 


Completing  the  square, a;- [--— = \"a~%' 

Ci  'iCL  CI  ~rfl/ 


Evolvmg, '  '-2-a=^4  \~^'^^r 

Transposing,  ^=2^=^  J  l^T +47^' 

Z>       1     f 
Reducing,  ^=^-±9-     [4a(c?— c)+5']. 

225.  If  the  equation  contains  two  powers  of  the  unknown 
quantity,  and  the  exponent  of  the  one  is  double  that  of  the 
other,  it  may  be  resolved  like  a  quadratic.     Thus, 

15.  Given  x'^-\-4:X^=117  to  find  the  values  of  .t. 

Conditions,  x^-\-Ax'^z=117. 

Completing  the  square,  a;^-}-42;'-}-4=117+4=121. 
Evolving,  a:'4-2=±ll. 

Transposing,  2;'^= ±11— 2=9,  or  —13. 

Evolving,  x=o,  or  a/ — 13. 

16.  Griven  x^ — 6a;^=16  to  find  the  values  of  2-. 
Conditions,  a;''— 6a;^=16. 
Completing  the  square,  x^—Qx^-\-d=lQ-\-9=i2b. 
Evolving,                       x'' — 3=zb5. 
Transposing,                  0:^=^54-3=8,  or  —2. 


Evolving,  x=2,  or  /^—2. 


17.  Given ^=22^  to  find  the  values  of  ^. 

J. 

Conditions,  ^ rr  =22^. 

2xi 
Clearing  of  fractions,     x ^  =44^. 


QUADRATIC  EQUATIONS.  179 


2x^      1  1      400 

Completing  the  square,  x — f--=44^-f--^— -. 


Evolving, 

4     1         20 

Transposing, 

J.         20     1     21     ^ 

19 
3 

Involving, 

a:=49,  or  +^-. 

18.  Given  3a;^"- 

-2:i'"=25  to  find  the  value  of  a:. 

Conditions, 

3:f2"— 2a;"=25. 

Dividing, 

,      2x''     25 
"-         3  -  8  • 

2a;"     1     25     1     76 
Completing  the  square,  x"" ^4--=— -f-=— . 

Evolving,  2;._-=__==___. 

1  2VT9   i+2vnr9 

iransposing,  5;"=-+ — - — == — —- . 

liVOlvmg,  2'^  I    — ! — J  "  . 

19.  Given  /i/4a:-|-10=12  to  find  the  value  ofx. 

Conditions,  /s/  4iX -{-1^=11. 

Squaring  both  sides  of  the  equation,  42:-|-16=144. 

Transposing,  4a;=144— 16=128. 

Dividing,  a;=:32. 


20.  Given  /y2a:4-3-f-4=7  to  find  the  value  of  a:. 


Conditions,  A^2a;4-3+4=7. 


Transposing,  /i^2z-{-3=7 — 4=3. 

Involving  both  sides,  22:-f-3=27. 


180  ALGEBRA. 

Transposing,  2xz=27 —S=24t. 

Dividing,  a:=12. 

21.  Given  V  V2-\-x=2-{-^/^to  find  the  value  of  x. 

Conditions,  a/  12-\-x=2-\-a/x'. 
Squaring  both  sides,  12-}-^=4-|-4/\/^4-a:. 

Transposing,  &e.,  8=:4V"£ 

Dividing,  2=a/x'. 

Involving,  4=a:. 

22.  Given  V^H" 40=10— V^  to  find  the  value  of  a:. 
Conditions,  a/  x -{-4:0=10— a/x. 
Squaring  both  sides,  re  4-40=100— 20/%/^ 4-a;. 
Transposing  and  reducing,  '       20/y^"i=60. 
Dividing,                                                  ,  a/'^=  ^• 
Involving,  xz=  9. 


23.  Given  /^x — a=/s/x — ^/s/a  to  find  the  value  of  x. 
Conditions,  a^  x—a=i/\/  x—^a/  a. 
Involving,  x—a=x—A/ax-{-j. 

Transposing,  Vtzz=flj-j"T=x* 

Involving,  ax=— Tr- 

io 

Dividing  by  a,  ^~T6  * 

.8 

24.  Given  Sx^ ^=_592  to  find  the  values  of  x. 

8 

Conditions,  So:^— -^  =  -592. 

8 

Changing  the  signs,  &c.,  — 3a:^=592. 

8     62;^     1184 
Multiplying  by  f ,  x^ =—^^, 


QUADRATIC  EQUATIONS.  181 


^       ,    .      ,,                    8     Gx^      9      1184  .   9      5929 
ComiDietmg  the  square,  X'' —-{-—= 


Extracting  the  root,  x^ — -=-1 — --. 


5     '  25        5     '  25""  25  * 
4     3         77 
5==^"5- 


4         77     3  74 

Transposing,  x'-^:=:±:-^-\-p=lQ,  or  — —. 

0         0  0 

/     74\3 
Evolving,  x=S,  or  f  — ~  J^. 

21 
25.  Given  V^^'+l+^/</^= —        —  to  find  the  values  of  x. 

a/2x-\-1 

_  21 

Conditions,  ^  2x -{-!-]- 2 /^/x 


V2a;+1 

Clearing  of  fractions,  2z+l+2V2^M^=21. 

Transposition,  2a/2x'-\-x=20—2x. 

Division,  a^  2x--{-x=10—x, 

Squaring  both  sides,  2x^-{-x=100—20x-\-x^. 

Transposing,  x--\-21x=  100. 

441  441     841 

Completing  the  squares,  x"-\-21x-\ — r— =100-] — -=— — . 

4  4         4 

21         29 
Evolution,  x-\-Y=do-^- 

29     21 

Transposition,  ^=±7^ — "o"^^'^'  ^^'  —25. 


26.  Given  2v'a;— fl;+3,v^±i;= — -^^^^^ —  to  find  the  values  of  a:. 

a/x — a 

'a-^5x 


Conditions,  2a/x — a-\-3/s/2x= 


As/  X — a 

Multiplying,  2x — 2a-{-S/y/2x^ — 2ax=7a-\-Dx. 

Transposing,  S/s/2x-^ — 2ax=da-{-ox 

Dividing,  a/2x-^ — 2ax=zoa-l-x. 

Involving,  2x' — 2ax=:da^-{-Qax-{-z^. 

Transposing,  x^ — Sax=9a^. 


18i2  A  L  a  E  B  R  A  . 

Completing  the  squares,        x^ —Sax -{-160^=250,^. 
Evolving,  x—4:a=-±:ba. 

Transposing,  .T=±5a-|-4a=9a,  or  —a. 

27.  Given  x-^^=/s/ x-{-b-\-Q  to  find  the  values  of  x. 

A71S.  x=4:,  or  — 1. 

28.  Given  A>/bx^\^=A/5x-\-2  to  find  the  value  of  x. 


Conditions,  A/bx  +  10=A^  bx-{-2. 
Squaring  both  sides,  bx-{-10=bx-{-4:Ay  bx-\-4:. 

Transposing,  &c.,  6=4/\/5^- 

Dividing,  S=2a/5x. 

Involving,  9=20a;. 

Dividing,  &c.,  a;=/^. 

29.  Given  ^^/^±3^^^/^_^  to  find  the  value  of  a:. 
/>/^-}-3      A/x^-\-^b 

s/lc-^la     sfx-^^a 

(Jonditions,  — -— = — -=. * 

fs/x-^-h      /sj  x-\-'^h 

Multiplying  both  sides  of  the  equation  by  s/lc-\-h  and  sfx-\-Zh^ 
we  have 

Reducing,  &c.,  (2a— 2^»)XA/V=2a^. 

ah 


Dividing,  /s/x= 


a — b' 


/  ab  \- 
Involving,  x=z\——^j. 

1     1        fl         I   4       y 
).  Given — I — =      -+    \-^r^A — -  to  find  the  value  of  a:. 


^     ,.  .  Ill, 

Conditions,  — I — = ^  —  4- 

x^  a      Sa^     \ 


2    2    I        4 


12      11         14       9 

Squaring  both  sides,      ~A 1 — :,=— +      -— t,H — r- 

■  ^  x^  ^  ax     a'     «-  '  \a-x^     x^ 


QUADllATIC 

EQUATIONS. 

Transposing,  &c., 

Multiplying  by  a:, 

Squaring  both  sides, 

L+i +1=1+1 

x^  ^  ax     (V-     d^     X' 

Reducing,  &c., 

4__8 
ax     X?' 

Dividing,  &c., 

1_2 

a     X 

Transposing,  &c., 

x=.2a. 

183 


31.  Griven  x=^ a^-\-X/^b^-{-x^—a  to  find  the  value  of  a:. 

Ans.  x= — -. . 

4a 

32.  Given  — -z — =^!^  to  find  the  value  of  a-. 

a/x         X 

Ans.  x= 


1—a 

83.  Given  a;2+12a:— 16=92  to  find  the  values  of  a:. 

A71S.  x=Q,  or  — 18. 

34.  Given  a:^— 3a;=10  to  find  the  values  of  2;. 

Am.  x=b,  or  — 2. 

35.  Given  a;^— a;4-3=45  to  find  the  values  of  a:. 

Ans.  x=7,  or  —6. 

36.  Given  bx'^-}-x=4:  to  find  the  values  of  x. 

Ans.  x=-=;  or  — 1. 
0 

37.  Given  2x'^ — a;=21  to  find  the  values  of  a:. 

Ans.  xJ-,  or  -3. 

38.  Given  5a;-+6a;— 3=60  to  find  the  values  of  a:. 

21 

Ans.  a:=o,  or  — ::-. 

0 

39.  Given  (a:— 12)(a;-|-2)=0  to  find  the  values  of  a:. 

Ans.  a:=12,  or  —2. 


184  ALGEBRA. 

40.  Given  S:^^— 14a;+15=0  to  find  the  values  of  x. 

Ans.  x=Z,  or  1|. 

41.  Given  ax^ — bx:=zc  to  find  the  values  of  a;. 

^,„.  ,^^±V(f+M. 

42.  Given  4:c-— 6:?;=108  to  find  the  values  of  x. 

Ans.   x=  6,  or  — 4^-. 

14— a; 

43.  Given  Ax — r-=14  to  find  the  values  of  x. 

x-\-l 

7 
Ans.  x=4:,  or  — -. 
4 
10     14 2x     22 

44.  Given — =~  to  find  the  values  of  x. 

X  X-  9 

o        21 

A71S.  2-=d,  or  ~. 


45.  Given  a;-|-^5:r-fl0=8  to  find  the  values  of  x. 

Ans.  a;=18,  or  3. 

46.  Given  x  -|-V10^-+6=9  to  find  the  values  of:?:. 

Ans.  2-=25,  or  3. 

47.  Given  3:i;2-f  22'— 9=76  to  find  the  value  ofx. 

Ans.  .^=5,  or  — 5§. 

48.  Given  x'—l^x=—2b  to  find  the  value  of  a:. 

Ans.  x=zb. 

49.  Given  32:'— a:— 140  =  0  to  find  the  value  of  a;. 

Alls.  x=7,  or  -2jf . 

7a: 

50.  Given  bx^~\——=:7x^ — 51  to  find  the  value  of  2:. 

Ans.  2:= 6,  or  — 54-. 

due  of  2'. 

Ans.  2'=:  4,  or  ^. 

52.  Given  ^+202:=32:2— 80  to  find  the  value  of  a:. 

An^s.  2:=:10,  or  — 2f . 

53.  If  2;'^-f-82:=65,  what  are  the  two  values  of  a:  1 

Ans.  2:=5,  or  — 13. 


42: — 4 
51.  Given  22;^ ^-— — =72;  to  find  the  value  of  2*. 


QUADRATIC  EQUATIONS.  186 

54.  If  6a;2— 2;=92,  what  are  the  two  values  of  a;  ? 

23 

Ans.  x=4:,  or  — ^. 

55.  If  dx^-\-4:X=M0,  what  are  the  two  values  of  a;  ? 

A?is.  a:=10,  or  — 11^. 

56.  If  a;2— 10a;=— 21,  what  are  the  two  values  o^x? 

A71S.  x=7,  or  3. 

57.  If  bx^—-=7S,  what  are  the  two  values  ofx? 

Ans.  x=4:,  or  — S^^j- 

58.  If  lla;2— 100a;=— 201,  what  are  the  two  values  ofx? 

Ans.  a:=3,  or  Qj\. 

59.  If  32;2— 17a;=2a;2+84,  what  are  the  two  values  oix? 

A71S.  a;=21,  or  —4. 

60.  Given  a:+16— 7VFR^=10— 4.V^^+TB   to  find   the 
values  of  x.  Ans.  2*=9,  or  —12. 

61.  Given   9a;-fVT6p+3BP=15x^— 4  to  find   the   values 

4  1 

of  a:.  Ans.  x=-,  or  —^. 

o  o 

124- 8a;^ 

62.  Given  x= — ^^^-  to  find  the  values  of  x.       -^ 

X — 5 

Ans.  a; =9,  or  4. 

63.  Given  /'a:2--'y'  +  ('a2-3Y=^  to  find  the  value  of  a:. 

Arts.  a;=dz'2      -. —  • 

2 

64.  Given  a:— 1=2H to  find  the  values  of  a:. 

x^ 

Ans.  a; =4,  or  1. 

65.  Given  /s/l?^^=x—h  to  find  the  values  of  x. 

b        Ha?-b' 
Ans.  x=- 


^2^^~12^ 


A/4a:+2     4— V^ 

66.  Given  -7- — ;=i= p—  to  find  the  values  of  x. 

4-|-Va;         sf^ 

Ans.. ^,,or'^. 

16*  {See  Key,  p.  119.) 


186  ALGEBKA. 

67.  Given  a/^ — 1/s/x—x=.^js/ x  to  find  the  values  of  a:. 

Ans.  a;=4,  or  1. 

68.  Given  /s/^-\-/\/x^=lQ/s/x  to  find  the  values  of  a:. 

Ans.  a;=2,  or  — 3. 

X  sJ  X 

69.  Given  -=22|-j — —  to  find  the  values  of  x. 

.o        S61 
Atis.  a;=4y,  or  — — . 

70.  Given  = ?\t=0  to  find  the  values  of  a:. 

Am.  a:=49,  or  25. 

71.  Given  x^-{-x'^=7t)Q  to  find  the  values  of  a:. 

Ans.  a:=243,  or  —28^. 

lies  of  a:. 

Ans.  a:=4,  or  /^4d 

15 

73.  Given  V5-|-^~I"V^=    /-        to  fi^^^i  t^^  value  of  a;. 

-4?w.  a:=4. 

74.  Given  /s/ x-{-V2-{-/s/ x-\-12=Q  to  find  the  values  of  x. 

A71S.  4,  or  69. 

n 

75.  Given  a;" — 2ax^=b  to  find  the  values  of  x. 

Am.  x=:  (a± V^"+^)''- 

8 

4.     5a^^ 

76.  Given  3a;^ ^=—592  to  find  the  values  of  a:. 


2_ 

72.  Given  x^ — a;^=56  to  find  the  values  of  a:. 


Ans.  x=S,  or  (  — p- J  . 


Problems. 

1.  A  merchant  bought  a  number  of  pieces  of  two  kinds  of 
silk,  for  £92  3^.  There  were  as  many  j^ieces  bought  of  each 
kind,  and  as  man}''  shillings  paid  per  yard  for  them,  as  a  piece 
of  that  kind  contained  yards.  Now,  two  pieces,  one  of  each 
kind,  together  measured  19  yards.  How  many  yards  were  there 
in  each  ? 


QUADRATIC     EQUATIONS.  187 

Let  X  =  the  number  of  yards  in  one  piece ;  it  -will  also 
equal  the  number  of  pieces,  and  also  the  number  of  shillings 
per  yard ;  and  19— a;  =  the  number  of  yards  in  the  other 
piece. 

Therefore,   ar^-|~(l^ — xf=i  the  value  of  both  kinds. 
And  a;3+(19— a:)''=1843. 

Or  57a;-— 1083a;4-6859=1843. 

By  transposition,    57a;-— 1083a:=— 5016. 
Or  a;2— 19z=-88. 

^       1    .       ,                          o    -in    ,  361     361     ^^     9 
Completing  the  square,         a;-— ly^c-j — j-  =—7 oo=t- 

19         3 

Evolution,  X —  -^=--±i-^. 

3,19     ,,       „ 
a:=±-+— =11  or  8. 

19_2;=:8  or  11. 
Both  values  answer  the  conditions  of  the  question ;   therefore 
there  were  11  yards  in  one,  and  8  in  the  other. 

2.  The  plate  of  a  looking-glass  is  18  inches  by  12,  and  is  to 
be  framed  with  a  frame  all  parts  of  which  are  of  equal  width, 
and  whose  area  is  to  be  equal  to  that  of  the  glass.  Bequired 
the  width  of  the  frame.  Ans.  3  inches. 

3.  A  grazier  bought  as  many  sheep  as  cost  him  £60,  out  of 
which  he  reserved  15,  and  sold  the  remainder  for  £54,  gaining 
two  shillings  a  head  on  them.  How  many  sheep  did  he  buy, 
and  what  was  the  price  of  each  ? 

Ans.  75  sheep,  at  16  shillings  each. 

4.  A  merchant  sold  a  quantity  of  flour  for  839,  and  gained 
as  much  per  cent,  as  the  flour  cost  him.  AYhat  was  the  price  of 
the  flour?  Aiis.  830. 

5.  There  are  two  numbers,  whose  difi"erence  is  9,  and  whose 
sum  multiplied  by  the  greater  is  266.  What  are  those  num- 
bers ?  Ans.  14  and  5. 

6.  A  and  B  gained,  by  trade  $18;    A's  money  was  in  the 


188  ALGEBRA. 

firm  12  months,  and  he  received,  for  his  principal  and  gain,  $26. 
B's  money,  which  was  $30,  was  in  the  firm  16  months.  What 
money  did  A  put  into  the  firm  ?  Ans.   $20. 

7.  A  merchant  bought  a  quantity  of  flour  for  $72,  and  he 
found  that  if  he  had  bought  6  barrels  more  for  the  same  money, 
he  would  have  paid  $1  less  for  each  barrel.  How  many  barrels 
did  he  buy,  and  what  was  the  price  of  each  ? 

Alls.  He  bought  18  barrels,  at  $4  per  barrel. 

8.  A  square  court-yard  has  a  gravel-walk  around  it.  The 
side  of  the  court  wants  2  yards  of  being  6  times  the  breadth  of 
the  gravel-walk,  and  the  number  of  square  yards  in  the  walk 
exceeds  the  number  of  yards  in  the  perimeter  of  the  court  by 
164  yards.     Kequired  the  area  of  the  court. 

Atis.  256  square  yards. 

9.  Given  —— — rQ=^  to  find  the  values  of  a:. 

Ans.  x=2,  or  ^. 

10.  Given  x^ — 2s^-^x=lS2  to  find  the  values  of  x. 

Ans.  x= ^ . 

11.  Given  9x-\-/s/l^x'^-{-'Si5x^=l^x^ — 4  to  find  the  values 
of  a:.  Ans.  x=^,  or  — i. 

12.  It  is  required  to  find  two  numbers,  the  first  of  which  may 
be  to  the  second  as  the  second  is  to  16,  and  the  sum  of  the 
squares  of  the  numbers  may  be  equal  to  225. 

A?is.    9  and  12 

QUADRATICS   WITH   TWO   OR   MORE   UNKNOWN   TERMS. 

1.  Given  x-{-y==10  \        ^    ,    , 

A    J  T  /?  (   to  find  the  values  of  x  and  y. 

And        xy=lb  )  ^ 

(1.)  First  equation,  a:-|-2/=10' 

(2.)  Second  equation,  xy==lQ. 

(3.)  Squaring  the  1st,  x'^-\-2xy-}-2f=100. 

(4.)  Multiplying  (2)  by  4,  4xy       =64. 


QUADRATIC     EQUATIONS. 


189 


(5.)  Subtracting  4th  from  3d, 

x^-2xy-^f=Z^. 

(6.)  Evolving  5th, 

a:— 2/=±6. 

(7.)  The  1st, 

x^y=\^. 

(8.)  Adding  Gth  and  7th, 

2a;=16,  or  4. 

(9.)  Subtracting  Gth  from  7th, 

22/=4,  or  16. 

(10.)  Dividing  the  8th  by  2, 

a:=8,  or  2. 

(11.)  Dividing  the  9th  by  2, 

7/=2,  or  8. 

Hence, 


a;=8  or  2,  and  2/=2  or  8. 


This  method  may  be  adopted  whenever  the  sum  and  product 
of  two  unknown  quantities  are  given. 


2.  Given  x — 2/=3 


And       xy 


2/=3    \ 
=10    ) 


to  find  the  values  of  x  and  y. 


(1- 

(2. 
(3. 

(4. 
(5. 
(6. 

a- 

(8. 

(9. 
(10. 
(11. 


First  condition. 
Second  condition. 
Squaring  1st, 
Multiplying  2d  by  4, 
Adding  3d  and  4th, 
Evolving  the  5  th, 
The  1st, 

Adding  6th  and  7th, 
Dividing  8th  by  2, 
Subtracting  7th  from  Gth, 
Dividing  10th  by  2, 


X — y=^. 
xy=\S). 
a;- — ^xy-{-y^-=.%. 

42:?/=40. 
a;2^22'^+?/-=49. 

x-y=?.. 

2:r=10,  or  -4. 

a;=5,  or  —2. 
22/=4,  or  —10. 

2/=2,  or  —5. 


Hence,  a:=5  or  — 2,  and  y=2  or  — 5. 

We  may  proceed  in  the  same  manner  whenever  the  difi"erence 
and  product  of  two  unknown  quantities  are  given. 

3.  Given  a;  +?/  =  20  ) 

And     or-X-  - '^OS  )  *°  ^^^  ^^  values  of  a;  and  y. 


(1.)  First  equation, 
(2.)  Second  equation, 
(3.)  2d  multiplied  by  2, 


.T+7/=20. 

:r+?/-=208. 
2a;24-22r'=416. 


190 


ALGEBRA. 


(4.)  Square  of  the  1st, 

(5.)  Subtracting  4th  from  3d, 

(6.)  Evolving  5th, 

(7.)  First  equation, 

(8.)  Sum  of  6th  and  7th, 

(9.)  Half  of  the  8th, 

(10.)  Subtracting  6th  from  7th, 

(11.)  Half  of  10th, 


a;2-j-2:i:?/4-7/2=400. 
x" — 2a:?/-j-?f=16. 
^~2/=±4. 
a;4-2/=20. 

2a:=24,  or  16. 
a:=12,  or  8. 
2y=16,  or  24. 
yz=  8,  or  12. 


Hence, 
4.  Given  x  — ?/  = 


x=\%  or  8;  2/=8,  or  12. 


And     x^-\-y 


3  ) 

2=117  ) 


to  find  the  values  of  x  and  y. 


(1- 

(2- 
(3. 

(4. 
(5. 
(6. 

(7. 

(8. 

(9. 
(10. 
(11. 


First  equation, 
Second  equation, 
The  2d  multiplied  by  2, 
Square  of  the  1st, 


x—y=.Z, 
2a;2+22/'=234. 


Subtracting  4th  from  3d,  x^-^lxy^y^'^ll^. 


Evolving  the  5th, 

The  1st, 

Sum  of  the  6th  and  7th, 

Dividing  8th  by  2, 

Subtracting  7th  from  6th, 

Dividing  10th  by  2, 


X — 2/  =  3. 

2:r=18,  or  —12. 

a:=9,  or  — 6. 
2?/=12,  or  —18. 

2/=6,  or  —9. 


Hence, 


a;=9,  or  —6 ;  2/=6,  or  —9. 


5.  Given  ^^^1^=10  )       .    ,  ^,        ,         .         , 

2 no  \  to  nnd  the  values  or  x  and  y. 


And         X- — y 


(1.)  First  equation, 
(2.)  Second  equation, 
(3.)  Square  of  the  1st, 
(4.)  Sum  of  2d  and  3d, 
(5.)  Half  the  4th, 
(6.)  Square  root  of  5th, 


V^:q:^=l0. 

a:2-2/2=28. 
a;2+2/2=100. 
2a;2=128. 
a:-=64. 
x=%. 


QUADRATIC     EQUATIONS.  191 

(7.)  Subtract  2d  from  3d,  2y'-=72. 

(8.)  Half  the  7th,  7f=^(). 

(9.)  Square  root  of  8th,  y=^' 
Hence,                    x=S,  and  y=Q. 


And     r'+2/'^=35  ) 

\J1.    >Aj    <xuva     u. 

(1.)  First  equation. 

a;+?/=5. 

(2.)  Second  equation, 

:^-^y'=Zh. 

(3.)  Square  of  the  1st, 

a^-\-2xy-j-f=25. 

(4.)  The  2d  divided  by  the  1st 

x"—xy-{-y-=l. 

(5.)  Subtracting  4th  from  3d, 

Sxy=lK 

(6.)  Dividing  5th  by  3, 

xy=6. 

(7.)  The  4th, 

x'^—xy-\-y-=l. 

(8.)  The  6th, 

xyz=Q. 

(9.)  Subtracting  6th  from  7th, 

x^—2xy-\-y-=\. 

(10.)  Evolving  the  9th, 

x—y=l. 

(11.)  The  1st, 

x-\-y=b. 

(12.)  Sum  of  10th  and  nth, 

2x=Q, 

(13.)  Half  of  12th, 

a:=3. 

(14.)  Subtracting  10th  from  11th, 

%=4. 

(15.)  Half  of  14th, 

y=2. 

Hence,                    a:  =  3,  and  2/= 2. 

7.  Given  x'^y'=.2^  \       .    .    . 

A    J       o       >     n  n  ( to  find  the  values  of  x  and  v. 
And     X- — y'=.\2  )  ^ 

Ans.  x=4:;  y=2. 

8.  Given  x  -\-y  =  6  ) 

.    ,       .,  .    o     r^r,  c  to  nnd  the  values  ot  x  and  v. 
And    x--{-y-=2Q  )  ^ 

Ans.  x=b,  and  y=l. 

9.  Given  x'^-i-f-=74:  K    .    ,  .        ,         .         , 

.     ,  ^  c  to  nnd  the  values  oi  x  and  ?/. 

And      x—y  =   2  )  ^ 

Atis.  x=i7,  and  2/= 5. 

A    1  ,  -,  ^  (  to  find  the  values  of  x  and  y. 

And     .r  -f?/  =  17  )  ^ 

^7W.  2:=10,  and  ?/=7. 


192  ALGEBRA. 

11.  Given  a:-— ?/-=85  K    ^    i  ^         i         r.         ^ 
.    ,  .         -, «.  (  to  nnd  the  values  of  x  and  y. 

And       x  +  y=17  )  ^ 

^715.  :i*=ll,  and  y=Q. 

]^2.  Given  2^ ?/  =2    ) 

'    .     T       o       <?     f^o  C  to  fij^*!  the  values  of  x  and  ?/. 
And    2'" — 2/'^= 9 8  )  *^ 

J.7Z5.  a:=5,  and  2/=3. 

13.  Given  10x-\-yz=Sxy  )       ^    ,    , 

•     1  r»      (  to  nnd  the  values  01  x  and  y. 

And        y—x=2      )  ^ 

Ans.  x=2  or  — -,  and  ?/=4  or  -J-k. 
3  *^  3 

EXAMPLES    OF    ONE   OR   MORE   UNKNOWN   TERMS. 

1.  A  says  to  B,  The  sum  of  our  money  is  18  dollars;  B  re- 
plies, But  if  twice  the  number  of  your  dollars  were  multiplied 
by  mine,  the  product  would  be  $154.  How  many  dollars  had 
each  ?  Ans.  A  had  $7,  and  B  had  $11. 

2.  The  difference  of  two  numbers  is  5,  and  the  sum  of  their 
squares  is  193.     "What  are  those  numbers  ?      A?is.  12  and  7. 

8.  A  and  B  have  each  a  small  field,  each  of  which  is  an 
exact  square,  and  it  requires  200  rods  of  fence  to  enclose  both. 
The  contents  of  these  fields  are  1300  square  rods.  AVhat  is  the 
value  of  each,  at  $2.25  per  square  rod? 

Ans.  A's  field,  $900 ;  B's,  $2025. 

4.  A  lady  wishes  to  purchase  a  carpet  for  each  of  her  square 
parlors,  one  of  which  is  3  feet  longer  than  the  other,  and  it  will 
require  85  square  yards  for  both  rooms.  Mr.  Ames  has  good 
carpeting,  which  is  40  inches  wide,  which  he  will  sell  at  $1.75 
per  yard.  What  will  it  cost  the  lady  to  carpet  each  of  her 
I'ooms  ?     A71S.  For  the  larger  room,  $77.17^ ;  smaller,  $56.70. 

5.  There  are  two  piles  of  wood,  each  of  which  is  a  perfect 
cube ;  the  sum  of  their  lengths  is  20  feet,  and  their  contents 
are  2240  cubic  feet.  What  is  the  value  of  each  pile,  at  $6.25 
per  cord? 

Am.  Value  of  the  larger  pile,  $84.37^-;  the  smaller,  $25. 

6.  There  are  two  square  buildings,  that  are  paved  with  stones 
a  foot  square  each.     The  perimeter  of  the  larger  building  ex- 


QUADRATIC  EQUATIONS.  193 

ceeds  that  of  the  smaller  by  48  feet,  and  both  their  pavements 
together  contain  2120  stones.  What  are  the  lengths  respect- 
ively ?  Am.  26  and  38  feet. 

7.  A  sets  out  from  Boston  for  Portland,  the  distance  being 
105  miles.  B  sets  out  at  the  same  time  from  Portland  for 
Boston.  A  arrives  in  Portland  in  9  hours,  B  arrives  in  Boston 
hi  16  hours,  after  they  meet.  In  what  time  does  each  perform 
the  journey  ?  Ans.  A  in  21  hours  ;  B  in  28  hours. 

8.  Divide  60  into  two  such  parts  that  their  product  shall  be 
to  the  difference  of  their  squares  as  2  to  3.     Aiis.  40  and  20. 

9.  There  are  two  numbers  whose  product  is  77,  and  the 
difference  of  whose  squares  is  to  the  square  of  their  difference 
as  9  to  2.     Required  the  numbers.  A7is.  11  and  7. 

10.  I  have  two  house-lots,  the  contents  of  which  are  225 
square  rods,  and  the  area  of  the  less  is  to  the  area  of  the  larger 
as  9  to  16.     Required  the  contents  of  each  lot. 

A71S.  81  square  rods  in  the  less,  and  144  in  the  larger. 

11.  The  product  of  two  numbers  is  48,  and  the  difference  of 
their  cubes  is  to  the  cube  of  their  difference  as  87  to  1.  Re- 
quired the  numbers.  Ans.  8  and  6. 

12.  There  are  two  numbers  whose  product  is  196,  and  if  the 
greater  be  divided  by  the  less  the  quotient  is  4.  What  are 
those  numbers  ?  A?is.  28  and  7. 

13.  A,  B  and  C,  can  perform  a  piece  of  work  in  a  certain 
time ;  A  can  perform  it  in  6  hours,  B  in  15  hours,  and  C  in  10 
hours.     How  long  would  it  take  them  all  to  perform  it  ? 

Ans.  3  hours. 

14.  A  grazier  bought  a  certain  number  of  oxen  for  8240,  and 
having  lost  3,  he  sold  the  remainder  at  $8  a  head  more  than 
they  cost  him,  thus  gaining  $59  by  his  bargain.  What  number 
did  he  buy  ?  Ans.  16. 

15.  The  paving  of  two  court-yards  cost  £205  ;  a  square  yard 
of  each  cost  }  as  many  shillings  as  there  were  yards  in  a  side 

17 


194  ALGEBRA. 

of  the  other ;  and  a  side  of  the  greater  and  less  together  measure 
41  yards.     Required  the  length  of  a  side  of  each. 

Ans.  25  and  16  yards. 

16.  Divide  145  into  two  such  parts,  that  the  sum  of  their 
square  roots  shall  be  17.  Atis.  81  and  64. 

17.  Sold  an  ox  for  $56,  and  gained  as  much  per  cent,  as  the 
ox  cost.     What  was  paid  for  him  ?  Atis.  $40. 

18.  Divide  the  number  14  into  two  parts,  so  that  the  sum  of 
their  cubes  shall  be  728.  Ans.  8  and  6. 

19.  My  farm  is  a  rectangle,  and  the  length  is  twice  its  breadth ; 
but,  having  enlarged  it  two  rods  on  all  sides,  I  find  its  contents 
increased  496  square  rods.  Of  how  many  acres  does  my  farm 
at  present  consist  ?  Ans.  23  acres,  16  rods. 

20.  There  are  two  numbers  whose  product  added  to  the  sum 
of  their  squares  is  109,  but  the  difference  of  whose  squares  is  24. 
Required  those  numbers.  Ans.  5  and  7. 

21.  What  number  is  that  to  which  if  40  be  added,  and  the 
square  root  extracted,  this  root  shall  be  less  than  the  original 
quantity  by  16  ?  Ans.  24. 

22.  Two  gentlemen,  A  and  B,  speaking  of  their  ages,  A  said 
that  the  product  of  their  ages  was  750.  B  replied,  that  if  his 
age  were  increased  7  years,  and  A's  were  lessened  2  years,  their 
product  would  be  851.     Required  their  ages. 

Ans.  A's  25  and  B's  30  years. 

23.  John  Smith's  garden  is  a  rectangle,  and  contains  15,000 
square  yards ;  and  he,  being  a  man  of  taste,  has  surrounded  it 
with  a  walk  7  yards  wide,  the  contents  of  which  are  3696  square 
yards.     Required  the  length  and  breadth  of  the  garden. 

Ans.  Length  150,  breadth  100  yards. 

24.  A  gentleman  purchased  a  farm  for  $5600,  but  if  his  farm 
had  contained  10  acres  more  it  would  have  cost  him  $10  less 
per  acre.     Of  how  many  acres  did  his  farm  consist  ? 

Ans.  70  acres. 

25.  A  man  purchased  a  farm  in  the  fonn  of  a  rectangle, 
whose  length  was  four  times  its  breadth.     It  cost  ^  as  many 


QUADRATIC     EQUATIONS.  195 

dollars  per  acre  as  the  field  was  rods  in  length,  and  the  number 
of  dollars  paid  for  the  farm  was  four  times  the  number  of  rods 
round  it.  Required  the  price  of  the  farm,  and  its  length  and 
breadth. 

Am.  Price  $1600.     Length  160  rods,  breadth  40  rods. 

26.  Two  men,  A  and  B,  set  out  from  the  same  place  at  the 
same  time  to  travel  to  Boston,  it  being  39  miles  distant.  A 
travelled  ^  of  a  mile  an  hour  faster  than  B,  and  arrived  at 
Boston  an  hour  sooner.     Required  the  rates  of  travelling. 

Ans.  A  3 J-  and  B  8  miles  per  hour. 

27.  What  two  numbers  are  those  whose  difference  multiplied 
by  the  less  produces  42,  and  by  their  sum  133  ? 

Ans.  13  and  6. 

28.  A  certain  company  agreed  to  build  a  vessel  for  $6300 ; 
but,  two  of  their  number  having  died,  those  that  survived  had 
each  to  advance  $200  more  than  they  otherwise  would  have 
done.     Of  how  many  persons  did  the  company  at  first  consist  ? 

Ans.  9  persons. 

29.  I  have  a  rectangular  field  of  corn,  which  consists  of  6250 
hills,  but  the  number  of  hills  in  the  length  exceeds  the  number 
in  the  breadth  by  75.  Of  how  many  hills  does  the  length  and 
breadth  consist  ?       Ans.  125  hills  the  length,  50  the  breadth. 

30.  A  man  bought  10  ducks  and  12  turkeys  for  $22.50.  He 
bought  4  more  ducks  for  $6  than  turkeys  for  $5.  What  was 
the  price  of  each  ? 

Ans.  The  price  of  a  duck  was  75  cents,  and  of  a  turkey 
$1.25. 

31.  What  number  is  that  to  which  if  24  be  added,  and  the 
square  root  of  the  sum  extracted,  this  root  shall  be  less  than  the 
original  quantity  by  18  ?  Ans.  25. 

32.  A  has  two  gardens,  each  of  which  is  an  exact  square. 
They  contain  208  square  rods.  It  requires  80  rods  of  fence  to 
enclose  both  gardens.     Required  the  contents  of  each. 

Ans.  144  square  rods  ;  64  square  rods. 

33.  A  has  two  square  gardens,  and  it  requires  80  rods  of 
fence   to   enclose  them.     The  larger  contains   80  square   rods 


196  ALGEBRA. 

more  than  tlie  other.     How  many  square  rods  do  both  gardens 
contain  ?  Ans.  208  square  rods. 

34.  A  has  two  square  gardens,  and  the  side  of  the  one  exceeds 
that  of  the  other  by  4  rods,  and  the  contents  of  both  are  208 
square  rods.  How  many  square  rods  does  the  larger  garden 
contain  more  than  the  smaller  ?  Ans.  80  square  rods. 

35.  I  have  two  blocks  of  marble  which  are  exact  cubes,  and 
whose  united  lengths  are  20  inches,  and  they  contain  2240  cubic 
inches.     Required  the  surface  of  each. 

Atis.  Larger,  864  inches ;  smaller,  384  inches. 

36.  A  merchant  sold  a  bale  of  cloth  for  $75,  and  gained  as 
much  per  cent,  as  the  cloth  cost  him.  What  was  the  price  of 
the  cloth  ?  ■  Ans.  $50. 

37.  There  are  two  numbers  whose  difference  is  12,  and  whose 
sum  multiplied  by  the  greater  is  560.   What  are  those  numbers  ? 

Ans.  20  and  8. 

38.  The  plate  of  a  looking-glass  is  36  inches  by  12  inches. 
It  is  to  be  framed  with  a  frame  all  parts  of  which  are  of  equal 
width,  whose  area  is  448  square  inches.  What  is  the  width  of 
the  frame  ?  Ans.  4  inches. 

39.  Divide  100  into  two  such  parts  that  the  sum  of  their 
square  roots  shall  be  14.  Ans.  64  and  36. 

40.  A  square  court-yard  has  a  rectangular  gravel-walk  around 
it.  The  side  of  the  court  wants  one  yard  of  being  six  times  the 
breadth  of  the  gravel-walk,  and  the  number  of  square  yards  in 
the  walk  exceeds  the  number  of  yards  in  the  perimeter  of  the 
court  by  340.  What  is  the  area  of  the  court  and  width  of  the 
walk? 

A71S.  Area  of  the  court,  529  square  yards ;  width  of  the 
walk,  4  yards. 

41.  A  merchant  bought  54  bushels  of  wheat,  and  a  certain 
quantity  of  barley.  For  the  former  he  gave  half  as  many 
shillings  per  bushel  as  there  were  bushels  of  barley,  and  for  the 
latter  4  shillings  per  bushel  less.     He  sold  the  mixture  at  10 


QUADRATIC     EQUATIONS.  197 

shillings  per  bushel,  and  lost  £28165.  by  his  bargain.  \VTiat 
was  the  price  of  the  barley  ? 

Ans.  36  bushels  of  barley,  at  14  shillings  per  bushel. 

42.  I  have  165^-  square  feet  of  plank,  3  inches  in  thickness, 
with  which  I  intend  to  make  a  cubical  box.  Required  its  con- 
tents in  cubic  feet.  A?is.  125  cubic  feet. 

43.  I  have  a  small  globule  of  glass,  one  inch  in  diameter. 
How  large  a  sphere  may  be  made  of  it,  if  the  glass  is  to  be  only 
2-V  of  an  inch  in  thickness,  taking  it  for  granted  that  all  spheres 
are  to  each  other  as  the  cubes  of  their  diameters  ? 

Ans.  Inside  diameter,  1.775-j-  inches;  whole  diameter, 
1.875-}-  inches. 

44.  John  Smith  has  two  cubical'  boxes,  whose  united  lengths 
in  the  clear  are  20  inches,  and  their  solid  contents  are  2240 
cubic  inches.     "What  is  the  difference  of  their  contents  ? 

Ans.  1216  cubic  inches. 

45.  I  have  two  house-lots,  which  contain  6100  square  feet, 
and  the  larger  contains  1100  square  feet  more  than  the  less. 
Required  their  dimensions.  Ans.  50  and  60  feet  square. 

46.  Two  men,  A  and  B,  bought  a  farm  of  200  acres,  for 
which  they  paid  $200  each.     On  dividing  the  land,  A  says  to 

B,  If  you  will  let  me  have  my  part  in  the  situation  which  I  shall 
choose,  you  shall  have  so  much  more  land  than  I  that  mine  shall 
cost  75  cents  per  acre  more  than  yours.  B  accepted  the  pro- 
posal. How  much  land  did  each  have,  and  what  was  the  price 
of  each  per  acre  ? 

Am.  A  had  81.866  acres,  at  $2.443+ ;  B  had  118.133-}- 
acres,  at  $1,693-}-. 

47.  A  and  B  engaged  to  reap  a  field  for  90  shillings.  A 
could  reap  it  in  9  days,  and  they  promised  to  complete  it  in  5 
days.     They  found,  however,  that  they  were  obliged  to  call  in 

C,  an  inferior  workman,  to  assist  them  the  last  two  days,  in  con- 
sequence of  which  B  received  3^.  9d.  less  than  he  otherwise  would 
have  done.     In  what  time  could  B  and  C  each  reap  the  field  ? 

A71S.  B  could  reap  the  field  in  15  days,  and  C  in  18  days. 
17# 


198  ALGEBRA. 

SECTION  XVIII. 

CUBIC  AND   HIGHER  EQUATIONS. 

Art.  226i  A  Cubic  Equation  is  one  in  which  the  highest 
power  of  the  unknown  quantity  is  the  third  power. 

As,  x^ — ax^-\-hx=c. 

227.  A  Biquadratic  is  an  equation  in  which  the  highest 
power  of  the  unknown  quantity  is  the  fourth  power. 

As,  a;* — ax^-^-hz^ — cx=d. 

228,  An  equation  of  the  fifth  degree  'is  one  in  which  the 
highest  power  of  the  unknown  quantity  is  the  fifth  power. 

As,  a;' — ax'^-\-bx'^ — cx^-{-dx=e. 

And  so  on,  for  all  other  higher  powers. 

There  are  many  particular  and  very  prolix  rules  given  for  the 
solution  of  the  above-mentioned  equations ;  but  they  all  may  be 
readily  solved  by  the  following  easy 

Rule.  1.  Find,  by  trial,  two  quantities  as  near  the  true  root 
as  convenient,  and  substitute  them  separately,  in  the  given  equa- 
tion, instead  of  the  unknoion  quantity,  and  find  how  much  the 
terins  collected  together,  according  to  their  signs  -\-  or  — ,  differ 
from  the  known  members  of  the  equation,  noting  whether  these 
errors  are  in  excess  or  deficiency. 

2.  Multiply  the  difference  of  the  two  quantities  found,  or  taken 
by  trial,  by  either  of  the  errors,  and  divide  the  product  by  the 
difference  of  the  errors  ivhen  they  are  alike,  but  by  their  mm 
when  they  are  unlike.  Or,  we  may  say,  as  the  difference  or  sum 
of  the  errors  is  to  the  difference  of  the  two  assumed  quantities, 
so  is  either  error  to  the  correction  of  its  supposed  quantity. 

3.  Add  the  quotient  last  found  to  the  quantity  belonging  to 
that  error  when  its  supposed  quantity  is  too  little,  but  subtract 
it  when  too  great,  and  the  result  will  give  the  true  root  nearly. 

4.  Take  this  root,  and  the  nearer  of  the  two  former,  or  any 


CUBIC     EQUATIONS.  199 

other  that  may  he  found  nearer ;  and,  by  jrroceeding  in  like 
manner  as  above,  a  root  will  be  obtained  nearer  than  before. 
Proceeding  in  the  same  manner,  we  may  obtain  the  answer  to 
any  degree  of  exactness  required. 

Note  1.  —  It  is  best  always  to  employ  two  assumed  quantities,  that 
shall  differ  from  each  other  only  by  unity  in  the  last  figure  on  the  right, 
because  then  the  diflFerence,  or  multiplier,  is  only  1.  It  is  also  best  to  use 
always  the  less  error  in  the  above  operation. 

Note  2. —  It  will  be  convenient,  also,  to  begin  with  a  single  figure  at 
first,  trying  several  single  figures,  till  there  be  found  the  two  nearest  the 
truth,  the  one  too  little,  and  the  other  too  great ;  and,  in  working  with 
them,  find  only  one  more  figure.  Then  substitute  this  corrected  result  in 
the  equation  for  the  unknown  letter  ;  and,  if  the  result  prove  too  little, 
substitute  also  the  number  next  greater  for  the  second  supposition  ;  but, 
if  the  former  prove  too  great,  then  take  the  next  less  number  for  the  second 
supposition  ;  and,  working  with  the  second  pair  of  errors,  continue  the  quo- 
tient only  so  far  as  to  have  the  corrected  number  to  four  places  of  figures. 
Then  repeat  the  same  process  again  with  this  last  corrected  number,  and 
the  next  greater  or  less,  as  the  case  may  require,  carrying  the  third  cor- 
rected number  to  eight  figures,  because  each  new  operation  commonly 
doubles  the  number  of  true  figures.  Proceed  in  this  manner  to  any  extent 
that  may  be  wanted. 

EXAMPLES. 

1.  Find  the  root  of  the  cubic  equation  3?-\-x'^-{-x=ilO^. 
We  see  that  x  lies  between  4  and  5.     We  assume,  therefore, 
4  and  5  as  the  two  values  of  x. 


FIRST  SUPPOSITION. 

4         = 

x 

= 

SECOND 

SUPPOSITION. 
5 

16 

= 

X? 

= 

25 

64 

= 

X^ 

sums 

^== 

125 

84 

155 

100 

but  should  be 

100 

— 16  errors  -j-bb 

Sum  of  the  errors,  55-j-16=71. 
Then,  71  :  1  :  :  16  :  .2. 
Hence,  a:=4-|-.2=4.2  nearly. 


200 


ALGEBRA. 


Again,  let  a;=4.2  and  4.3. 


/ 


FIRST  SUPPOSITION. 
4.2 

17.64 

74.088 


95.928 
100 

—4.072 


X 

a? 

sums 


SECOND   SUPPOSITION. 

4.3 
18.49 
79.507 


102.297 
100 

+2.297 


Sum  of  the  errors,  4.072+2.297=6.369. 
As  6.369  :  .1  :  :  2.297  :  0.036. 
Hence  a;=4.3— .036=4.264  nearly. 
Again,  let  a:=4.264  and  4.265. 


FIRST  SUPPOSITION. 

4.264 
18.181696 
77.526752 

99.972448 
100 


SECOND   SUPPOSITION. 

4.265 
18.190225 
77.581310 


100.036535 
100 


0.027552  0.036535 

Sum  of  the  errors,  .027552+.036535=.064087. 
As  .064087  :  .001  :  :  .027552  :  0.0004299. 
Hence,  a;=4.264+.0004299=4.2644299  nearly. 

2.  Find  the  root  of  the  equation  x^ — 15a;^+63a;:=50. 
Here  it  is  evident  that  the  root  is  more  than  1.     We  then 
assume  the  two  values  of  a:  to  be  1.0  and  1.1. 
Then  63.0         =         63:r         =         69.3 

-15  =     —Ibx"  —18.15 

1  =  a;^  1.331 


49 
50 


sums 


— 1  errors 

Sum  of  the  errors,  1+2.481=3.481. 


52.481 
50 

+2.481 


CUBIC     EQUATI  ON  S. 


201 


As  3.481 

:  .1  :  :  1  :  .03 

Add       1.00 

Hence  x 

=         1.03 

nearly. 

gain,  let  x 

=  1.03  and  1.02 

Then 

64.89 

632; 

64.26 

- 

-15.9135         - 

•15:c2 

-15.6060 

1.092727 

sums 

1.061208 

50.069227 

49.715208 

50 

errors 

50 

-I-.069227 

-.284792 

.284792 

.354019  :  .01  :  :  .069227  :  .0019555. 
Hence  :i;=1.03-.0019555=1.02804  nearly. 

3.  Find  the  value  of  a;  in  the  equation  a;^-j-10:c--f-5a:=:260. 

Ans.  a:=4.1179857. 

4.  Find  the  value  ofx  in  the  equation  x'^—2x=b0. 

Am.  a;=3.8648854. 

5.  Find  the  value  of  re  in  the  equation  a;''— 3a;^— 75:c=10000. 

Ans.  :r=:10.2609. 

6.  Find  the  value  of  x  in  the  equation  x^-\-2x'^-{-Sx^-{-4:x'^-{- 
5.T=54321.  Ans.  a:=8.414455.* 

7.  I  have  a  cubical  block  of  marble,  and  if  the  superficial 
contents  were  added  to  its  solid  contents,  the  sum  would  be  432 
feet.     What  is  the  length  of  the  block  ?  A?is.  6  feet. 

8.  Five  times  the  cube  of  a  certain  number  exceeds  ten  times 
its  square  by  45.     Eequired  the  number.  Ans.  3. 

9.  The  fourth  power  of  a  certain  number  exceeds  ten  times 
its  square  by  375.     Required  the  number.  Am.  5. 


202  A  L  G  E  B  K  A  . 

SECTION  XIX. 

RATIOS. 

Art.  229 •  Ratio  is  tlie  relation  whicli  one  quantity  bears 
to  another  of  a  similar  kind,  with  respect  to  its  magnitude. 

230.  The  magnitude  or  value  of  a  ratio  is  estimated  by 
stating  how  often  one  quantity  or  number  contains  or  is  con- 
tained in  another.  Thus,  in  comparing  16  with  2,  we  observe 
that  it  has  a  certain  relative  magnitude  with  respect  to  2,  which 
it  contains  8  times ;  and,  if  we  compare  16  with  4,  we  observe 
that  it  has  a  different  relative  magnitude,  for  it  contains  4  only 
4  times.  Hence,  16  is  less  relatively,  when  compared  with  4, 
than  it  is  when  compared  with  2. 

231.  The  general  method  of  expressing  the  ratio  which  one 
quantity  or  number  bears  to  another  is  by  placing  two  points 
between  them.     Thus, 

The  ratio  of        12  to  4  is  expressed  by  12  :  4. 

19  to  9  "  "        by  19  :  9. 

«  aiob  "  "         by    a  :  b. 

232.  The  first  term  of  a  ratio  is  called  the  Antecedent,  and 
the  last  term  the  Consequent.  The  antecedents  in  the  preceding 
ratios  are,  therefore,  12,  19,  and  a,  and  the  consequents  4,  9, 
and  b. 

233.  Ratios  may,  therefore,  be  represented  in  the  form  of 
fractions,  by  making  the  antecedents  the  numerators,  and  the 
consequents  the  denominators ;  thus, 

12   19       ^  a 

express  the  ratios  of  12  to  4,  of  19  to  9,  and  of  a  to  b. 

234.  A  ratio  is  said  to  be  of  equality  when  the  antecedent 
is  equal  to  the  consequent. 

Thus  the  ratio  of  12  :  12,  or  of  a  :  a,  is  a  ratio  of  equality. 


RATIOS.  203 

235.  A  ratio  is  of  greater  inequality  when  the  antecedent  is 
greater  than  the  consequent.     Thus, 

The  ratio  of  a-\-b  :  c,  or  of  12  :  6,  is  a  ratio  of  greater 
inequality. 

236.  A  ratio  of  less  inequality  is  when  the  antecedent  is  less 
than  the  consequent.     Thus, 

The  ratio  of  a  :  a-\-b,  or  of  6  :  12,  is  a  ratio  of  less  ine- 
quality. 

Note.  —  It  is  evident  that  the  ratio  of  equality  may  always  be  repre- 
sented by  unity. 

COMPARISON   BY   RATIOS. 

237.  If  the  terms  of  a  ratio  are  both  multiplied  or  both 
divided  by  the  same  quantity,  the  value  of  the  ratio  is  not 
altered. 

The  ratio  of  a  :  Z»  is  expressed  by  the  fraction  -.     Let  both 

TlCl 

terms  of  this  fraction  be  multiplied  by  n,  and  it  becomes  -— . 

710 

4 
The  ratio  of  4  :  3  is  expressed  by  the  fraction  - ;  and,  if  the 

o 

12     4 

terms  of  this  fraction  be  multiplied  by  3,  it  becomes  — =-. 

Now,  since  the  value  of  a  fraction  is  not  altered  by  multiplying 

both  the   numerator  and  denominator  by  the    same    quantity, 

a     na  .  7-1  i  • 

-=— -,  or  the  ratio  a  :  0  \s  the  same  as  the  ratio  na  :  ?io,  and 

b     nb 

the  ratio  of  12  :  9  is  the  same  as  4  :  3.     Thus  the  ratio  of 

16  :  12,  both  terms  being  divided  by  4,  is  the  same  as  4  :  3. 

The  ratio  of  5  :  7,  both  terms  being  multiplied  by  3,  is  the 

same  as  the  ratio  of  15  :  21.     And  the  ratio  of  (v-  :  ab^  by 

dividing  by  a,  is  the  same  as  the  ratio  of  a  :  b. 

238.  Ratios  are  compared  together  by  reducing  the  fractions 
which  represent  them  to  a  common  denominator. 

Thus  the  ratios  of  7  :  9  and  10  :  13  are  represented  by  the 

7         10  91  90 

fractions  ^  and  — »  which  are  equivalent  to  r-—  and  — —  ;  and 


204  ALGEBRA 

91     .  90  . 

since  — —  is  greater  than  —— ,  we  infer  that  the  ratio  of  7  :  9 
is  greater  than  10  :  13. 

239.  When  the  antecedents  or  consequents  are  the  same  in 
two  or  more  ratios,  we  immediately  compare  those  ratios  to- 

17 

gether  by  expressing  them  in  a  fractional  form.     Thus,  since  -^ 

17 
is  greater  than  -^-,  the  ratio  of  17  :  5  is  greater  than  17  :  9 ; 

and,  since  — — ,  is  less  than  -,  the  ratio  of  a  :  a-\-b  is  less  than 
a-f-o  h  ' 

a  :  h. 

240.  A  ratio  of  greater  inequality  is  diminished,  and  a  ratio 
of  less  inequality  is  increased,  by  adding  the  same  quantity  to 
both  terms. 

Let  —  represent  any  ratio,  and  add  n  to  each  of  the  terms, 

.„  ,      a       ,  a-\-n     ...  .     , 

then  these  two  ratios  will  be  -  and  — -^ — ,  which  are  equivalent 

b  o-\-n 

obA-an        ,  ab-\-hn      .^        -n      ,  .       .i       ,     ^  . 

to  T-T^ — r  and  -r—-'-, — :.     JNow,  it  a  be  greater  than-^,  -  is  a 

„         ,      .  ,.  ,  ab-\-an  .  .        ab-\-b7i 

ratio  or  greater  inequality,  and  — — -; — r  is  greater  than  -— — — - , 
&  ^        •^'  b{b-\-n)      ^  b{b-^n)' 

therefore  -  is  diminished  by  adding  n  to  each  of  the  terms. 
b  JO 

But,  if  a  be  less  than  b,  then  t  is  a  ratio  of  less  inequality,  and 

ab-\-an  .    .       ^,        ab-\-bn     ^,        p        a    .     .  j   i       ., 

—— -! — r  IS  less  than  -—r — :  '■>  thereiore,  —  is   increased  by  the 
b{b-^n)  b{b-]-n)  '  b  ^ 

addition  of  n  to  both  terms. 


COMPOUND   RATIOS. 

241 .  Ratios  are  compounded  by  multiplying  their  antecedents 
together  to  form  a  new  antecedent,  and  their  consequents  to 
form  a  new  consequent.  The  resulting  ratio  is  called  the  su7n 
of  the  compounding  ratios. 


RATIOS.  205 

Thus,  the  ratio  of  «  :  ^  is  compounded  with  the  ratio  of  c  :  d 
by  multiplying  the  antecedents  a  and  c  together  for  a  new  ante- 
cedent, and  the  consequents  b  and  d  together  for  a  new  conse- 
quent, and  the  resulting  ratio  ac  :  bd  is  the  sum  of  the  com- 
pounding ratios  a  :  b  and  c  :  d. 

If  the  ratios  4  :  7,  6  :  11,  and  7  :  9  are  compounded  to- 
gether, the  resulting  ratio  is  4x^X7  :  7X11X9,  or  168  :  G93, 
which,  reduced  to  its  lowest  terms  by  dividing  both  terms  by 
21,  becomes  the  ratio  8  :  33. 

242»  When  any  ratio,  a  :  b,  is  compounded  with  itself  twice, 
thrice,  or  any  number  of  times,  denoted  by  n,  then  the  resulting 
ratios  are  a^  :  b^,  a"  :  W^  d^  :  Z**,  &c.,  and  are  called  the  dupli- 
cate, triplicate,  quadruplicate,  &c.,  ratios  of  the  primitive. 

As  the  indices  or  exponents,  2,  3,  and  ?z,  express  the  number 
of  times  the  ratio  of  a  :  ^  is  compounded  of  itself,  they  are 
called  the  measures  of  these  ratios. 

Since  the  index  may  be  any  quantity,  either  integral  or  frac- 
tional, let  the  fraction  be  -,  -,  -,  — ,  &c. ;  then, 

2   d   4  m 


The  ratio  of    a- 

it  cc        ^T 

1 

"  "     am 


b~  is  the  square  root  of  the  ratio  of  a  :  ^. 


i^  is  the  cube  root  of        "    "      o1a:b. 


1 


b^  is  the  fourth  root  of     "    "      of  a  :  b. 
b~i  is  the  mih.  root  of  "    "      of  a  :  b. 


243.  The  ratios  of  a^  :  b^,  a^  :  b^,  a^  :  b'^,  &c.,  are  also 
called  the  subduplicate,  subtriplicate,  subquadruplicate,  &c., 
ratios  of  a  to  b. 

TKOPORTION. 

214.  Proportion  consists  in  the  equality  of  ratios. 

a     c 
Thus,  if  the  ratio  of  o^  :  ^  is  equal  to  that  of  c  :  ^,  or  j=-j^ 

then  a,  b,  c,  d,  are  said  to  be  proportional.     The  numbers  3,  12, 

.       .     o      o      1        T   4      1 
4,  16,  are  proportionals,  for  --=zj,  and  — =-. 

IS 


206  ALGEBRA. 

This  equality  of  ratios  is  expressed  by  writing  the  four  quan- 
tities thus,  <z  :  3  :  :  c  :  6?,  and  is  read,  a  is  to  3  as  c  to  d. 

245.  In  algebraic  investigations  the  quantities  are  generally 

a     c 
expressed  like  fractions,  thus  -r=^. 

0     a 

Cb  C 

In  the  proportion  a  :  b  :  :  c  '  d,  or  -=-,  a  and  d  are  the 

o     a 

extremes,  and  b  and  c  the  means.     The  first  term  is  also  called 

the  first  antecedent,  and  the  second  the  first  consequent,  the 

third  term  the  second  antecedent,  and  the  fourth  term  the  second 

consequent. 

246.  If  in  a  series  of  proportional  quantities  each  consequent 
is  identical  with  the  next  antecedent,  these  quantities  are  said 
to  be  in  continued  proportion.  Thus,  a  :  b  :  :  b  :  c  :  :  c  :  d 
:  :  d  :  e  :  :  e  :  f,  &c.,  the  quantities  a,  b,  c,  d,  e,  f,  &c.,  are 
said  to  be  in  continued  proportion. 

247.  When  the  second  and  third  terms  are  identical,  as  in 
the  proportion  a  :  b  :  :  b  :  c,  then  b  is  said  to  be  a  mean  pro- 
portional between  the  extremes  a  and  c,  and  c  is  called  the  third 
proportional  to  a  and  b. 

248 e  The  product  of  any  number  of  ratios,  of  which  the  con- 
sequent of  each  ratio  is  the  antecedent  of  the  succeeding  one,  is 
the  ratio  of  the  first  antecedent  to  the  last  consequent. 

Let  the  ratios  he  a  :  b,  b  :  c,  c  :  d,  d  :  e,  e  :  f,  then  the  re- 
sulting ratio  is  aX^XcXdXe  :  bXcXdX^Xf,  or  the  ratio  of 
abode  :  bcdef,  which,  reduced  to  its  least  terms  by  cancelling 
the  same  letters  in  each  term,  becomes  a  :  /,  or  the  first  ante- 
cedent and  the  last  consequent. 

Again;  let  the  ratios  be  2  :  3,  3  :  4,  4  :  5,  5  :  7,  7  :  10, 
then  the  resulting  ratio  is, 

2X3X4X5X7  :  3x4x5x7x10,  or  840  :  4200, 
which  reduced  is,  7  :  35,  or  1  :  5. 

249.  Any  ratio  compounded  with  a  ratio  of  greater  inequality 
is  increased,  and  compounded  with  a  ratio  of  less  inequality  is 
diminished. 


RATIOS.  207 

Let  a-{-b  :  a  represent  the  ratio  of  greater  inequality,  and 
a  :  a-{-b  of  less  inequality.  Then  the  ratio  of  a-\-b  :  a,  com- 
pounded with  that  of  c  :  d,  gives  ac-{-bc  :  ad,  which  is  evidently 
greater  than  the  ratio  of  c  :  d;  and  the  ratio  of  a  :  a~\-b, 
compounded  with  that  of  c  :  d,  gives  ac  :  ad-\-bd,  which  is 
evidently  less  than  the  ratio  of  c  :  d. 

Ilcnce  the  ratio  of  c  :  ^  is  increased  by  compounding  it  with 
the  ratio  of  a-\-b  :  a,  and  diminished  by  comj^ounding  it^  with 
the  ratio  of  a  :  a-\-b. 

APPROXIMATION    OF    RATIOS. 

250i  The  ratio  of  the  powers  or  roots  of  two  quantities  whose 
difference  is  small  with  respect  to  themselves  is  found  very 
nearly  by  multiplying  that  difference  by  the  index  or  exponent 
of  the  power  or  root. 

PROPOSITIONS. 

Proposition  I.  If  four  quantities  are  proportional,  the  pro- 
duct of  the  extremes  is  equal  to  the  product  of  the  means,  and 
conversely. 

Let  a  '.  b  :  :  c  '.  d,  ov  y=^. 

b     d 

Multiplying  both  by  bd,  we  obtain  adz=bc. 

Conversely.  If  the  product  of  any  two  quantities  is  equal  to 
the  product  of  any  other  two,  these  four  quantities  are  propor- 
tional, the  factors  of  either  of  the  products  being  made  the 
extremes,  and  the  factors  of  the  other  the  means. 

d       C  C        Q, 

Let  ad=bc,  dividing  both  by  bd,  we  obtain  -j=-,,  or  -=- ; 

b     d        d     b 

whence  a  '.  b  :  i  c  :  d,  oi  c  :  d  :  :  a  :  b. 

Prop.  II.  If  three  quantities  are  in  continued  proportion, 
the  product  of  the  extremes  is  equal  to  the  square  of  the  mean, 
and  conversely. 

Let  a'.b'.'.b'.c;  aXc=^X^>  or  ac=b'^. 

Conversely.     If  the  product  of  any  two  quantities  is  equal  to 


208  ALGEBRA. 

the  square  of  a  third,  the  third  is  a  mean  proportional  between 
the  other  two. 

7 

Let  acz=.lr\    and,  dividing  both  by  ^c,  we  obtain  -=-,  or 

h     c 

a  :  b  :  :  b  :  c. 

Prop.  III.   Of  four  proportionals,  any  three  being  given,  the 
fourth  may  be  found. 

Let  a  :  b  :  '.  c  '.  d ;  then  ad=bc. 

T-T  i'C      ,      ad  ad      .     be 

Hence,  a=— -;  b= — ;  c=— -;  d=—. 

a  c  b  a 

Hence,   of  three   proportionals,    any   two   being   given,    the 

third  may  be  found;  for  ad=.b'^^  therefore  b=/sfad^  «=^-j  and 

d 

a 
Proi*.  IV.  Quantities  which  have  the  same  ratio  to  the  same 
quantity  are  equal  to  one  another,  and  conversely. 

a     c 

Let  a  :  b  :  :  c  :  b,  then  t=t  l  ^i^d,  multiplying  each  by  by 

we  obtain  a=c. 

Conversely.     Quantities  which  are  equal  to  one  another  have 
the  same  ratio  to  the  same  quantity. 

Let  a=:c,  and  let  3  be  a  third  quantity;  then,  dividing  both 
by  b,  we  obtain 

■5-=T'  therefore  a  :  b  :  :  c  :  b. 

0        0 

Prop.  N .   Ratios  that  are  equal  to  the  same  ratio  are  equal 
to  one  another. 

Let  a  '.  b  :  :  e  :  f,  and  c  :  d  '.  :  e  :  f;  then,  also,  a  :  b  :  :  c  '.  d. 

(L        S  C        S  CL        C 

Since  -=7;,  and  -,=-7,  it  is  evident  -7=-^j  ^"^^  therefore  a  :  b 
of  a    J  b     d 

:  :  c  '.  d. 

Or  let  2  :  4  :  :  8  :  16,  and  3  :  6  :  :  8  :  16. 

Then  2  :  4  :  :  3  :  6 ;  for  -,~  and  ?=^. 

4     2  0     2 

Prop.  VI.   Iffour  quantities  are  proportionals,  they  will  also 


RATIOS.  209 

be  proportionals  when  taken  inversely  ;  that  is,  the  second  will 
have  the  same  ratio  to  the  first  that  the  fourth  has  to  the  third. 

Let  a  '.  b  '.  :  c  :  dy  then  b  :  a  :  :  d  :  c. 

Since  by  Prop.  I.  bc=ad, 

And,  dividing  by  ac,  we  obtain    -=-, 

a     c 

Hence,  b  :  a  :  :  d  :  c. 

Prop.  VII.  If  four  quantities  are  proportionals,  they  will 
also  be  proportionals  when  taken  alternately ;  that  is,  the  first 
wiU  have  the  same  ratio  to  the  third  that  the  second  has  to  the 
fourth. 

Let  a  :  b  :  :  c  :  d  ;  then,  also,  a  :  c  :  :  b  :  d  ; 

As  j=-ji    if  we   multiply  each  quantity  by  -,    we    obtain 

ah      cb         ,  .  ,         -       .    .    c     3     ,        _  ,      , 

7— =—7-;  which,  reduced,  is  -=-,  therefore  a  :  c  :  :  b  :  a. 
be      cd  c      a 

This  may  be  illustrated  by  numbers ;  thus, 

Let  2  :  4  :  :  3  :  6,  then  2  :  3  :  :  4  :  6 ; 

2     3.  .  .  .4 

As  2;=pj  if  we  multiply  each  side  of  the  equation  by  •^,  the 

2434     8      12    24 
result  will  be  ^X^=gXg;  j^— 18'  s'^G'     ^^^^^^'^^^    ^-^ 
:  :  4  :  6. 

Prop.  VIII.  If  four  quantities  are  proportionals,  they  will 
also  be  proportionals  when  taken  jointly  ;  that  is,  the  sum  of  the 
first  and  second  ^ill  have  the  same  ratio  to  the  second  that  the 
sum  of  the  third  and  fourth  has  to  the  fourth. 

Let  a  '.  b  :  '.  c  :  d,  then  a-\-b  :  b  :  :  c-\-d  :  d. 

d       C  CL  C 

Since  -7= -j,  we  add  1  to  each  quantity,  and  obtain  -4-1=- 
0     d  b  a 

,  -I         (Z-\-b      c-{-d     ,0  ,     ,  ,77 

4-1,  or  — - — = — -— ,  thereiore  a  +c»  :  b  :  :  c-f-d  :  d. 
b  d 

This,  also,  may  be  made  evident  by  taking  numbers  ;  thus, 

Let  2  :  4  :  :  3  :  6,  then  2+4  :  4  :  :  3+6  :  6. 

18# 


210  ALGEBRA. 

2     3 
Since  ■7=tt>     we     add     1     to     each    number,    and    obtain 
4      D 

2  ,  ,      3  ,  _       2+4     3+6 
-+1=^+1,  or -^=^-. 

Therefore,  2+4  :  4  :  :  3+6  :  6. 

Prop.  IX.  If  four  quantities  are  proportionals,  they  will  also 
be  proportionals  by  separation  ;  that  is,  the  difference  between 
the  first  and  second  will  have  the  same  ratio  to  the  second  that 
the  difference  between  the  third  and  fourth  has  to  the  fourth. 

Let         a  :  b  :  :  c  :  d,  then  a — b  :  b  :  :  c — d  :  d. 

a     c 
Since  -7=3,  we  will  subtract  1  from  each  quantity,  and  we 
o     a 

,  ,  .  a     -     c     -        a — b      c — d 

obtam  - — 1=-^ — 1,  or  — ; —  :  — 7—. 

b  d  b  d 

Therefore,  a — b  :  b  :  :  c — d  :  d. 

This  demonstration  may  be  illustrated  by  numbers  ;  thus, 

Let         4  :  2  :  :  6  :  3,  then  4—2  :  2  :  :  6—3  :  3. 

4     6 
Since  75=0?    we    subtract    1    from  each  term,  and  we  have 

4    -,_6  4-2_6-3 

2  3       '°^     2    "~~3~* 

Therefore,  4—2  :  2  :  :  6—3  :  3. 

Prop.  X.  If  four  quantities  are  proportionals,  they  will  also 
be  proportionals  by  conversion  ;  that  is,  the  first  term  will  have 
the  same  ratio  to  the  sum  or  difference  of  the  first  and  second, 
that  the  third  has  to  the  sum  or  difference  of  the  third  and 
fourth. 

Let   a  :  b  :  '.  c  :  d ;    then   a  :  a-±Jj  :  c  :  :  c+ic?.      Since 

-=-,  and,  by  Prop.  VIII.  and  IX.,  — = — = — r—,  invert  these 
b     d  b  d 

fractions,  and  we  have  r= -',  and,  by  multiplying  the 

a-^^o     c-\~ct 

one  by  -,  and.  the  other  by  its  equal  -,  we  obtain  --Xt= 

b  a  a-hb      b 

d        c            a            c       ,,       „  .  ,  , 

,X  jj  or  r-= 3,  thereiore  a  :  <z±^  :  :  c  :  c-±:d. 


c-\-d     d        a-±J)      c-\-d 


RATIOS.  211 

Let  the  pupil  prove  this  by  numbers. 

Prop.  XI.  If  four  quantities  are  proportionals,  the  sum  of 
the  first  and  second  has  the  same  ratio  to  their  difference  that 
the  sum  of  the  third  and  fourth  has  to  their  difference. 

Let  a  :  b  '.  '.  c  :  d ;  then,  also,  a-\-h  :  a — b  :  :  c-{-d  :  c — d. 

For,  by  Prop.  VIII.  and  IX.  by  alternation,  a-\-b  :  c-]-d  : 
b  '.  d ;  and  a — b  :  c — d  :  :  b  :  d ;  hence,  by  Prop.  V.,  a-\-b 
c-\-d  :  :  a — b  :  c — d^  and,  by  alternation,  a-\-b  :  a — b  : 
c-\-d  :  c — d. 

This  is  illustrated  by  numbers,  thus ;  let  8  :  6  :  :  12  :  9 ; 
then  8+6  :  8—6  :  :  12-f  9  :  12-9. 

For  taking  Prop.  VIIL  and  IX.  by  alternation,  8-{-6  :  12 
+9  :  :  6  :  9;  and  by  Prop.  Y.,  8+6  :  12+9  :  :  8-6  :  12 
—9;  therefore,  by  alternation,  8+6  :  8—6  :  :  12+9  :  12 
-9. 

Prop.  XII.  In  any  number  of  proportionals,  any  antecedent 
has  the  same  ratio  to  its  consequent  that  the  sum  of  all  the 
antecedents  has  to  the  sum  of  all  the  consequents. 

Let  a  :  b  '.  :  c  :  d  :  :  e  :  f :  '.  g  :  h  ;  then,  also,  a  :  b  :  :  a-\-c 
+e+^  :  b-\-d-\-f-^h. 

Since  ab=ba,  ad=bc,  af=be,  ah-=.bg,  we  have 
a{b+d+f+h)=b{a-\-c+e-\-g) ; 

Whence,  -=    ,    ,  ,   ..  ,-r- ; 

b     ^+6?+/+/i 

Therefore,  a  :  b  :  :  a-\-c-\-e-{-g  :  ^+^+/+A. 

In  like  manner  it  may  be  shown  that  c  :  d  :  :  <z+c+c+o- 

This  proposition  may  be  illustrated  by  numbers,  thus. 

Let  2  :  3  :  :  4  :  6  :  :  8  :  12  :  :  14  :  21 ; 

Then  2:3::  2+4+8+14  :  3+6+12+21=2  :  3  : :  28  :  42. 

/^Prop.  XIII.     In  two  or  more     sets     of  proportionals,    the 
product  of  the  correspondent  terms  are  also  proportionals. 


212 


ALGEBRA. 


Let  a  :  b  :  :  c  '.  d,  \ 

e  :  f  '.  :  g  :  hj\  Then,  also,  aei  :  Ifk  :  :  cgl  :  dhm. 
i  :  k  :  :  I  :  m,) 


a     c    e     g  I 
bmce  T=^>  :?=T»  T== 


DEMONSTKATION. 

I      aXeXi       cXgXl 


h     d' f     Ilk     m'  bXfXk     dXhX^ 

Wlience  —-=z-^—,  therefore,  aei  :  hfk  :  cgl  :  :  dhm. 

hjk     dhm  q.  e.  d. 

ILLUSTRATION   BY   NUMBERS. 


6x10x14. 


Prop.  XIY.  If  there  are  any  number  of  quantities  more  than 
two,  and  as  many  others,  which,  taken  two  and  two  in  order,  are 
proportionals,  then,  by  equality,  are  the  extreme  terms  in  the 
former  series  proportional  to  the  extreme  terms  in  the  latter. 

Let  «,  ^,  c,  d^  be  any  number  of  quantities,  and  let  e,  /,  g^  A, 
be  as  many  others. 


Let 

2 

:  3  :  : 

4 

:     6 

4 

:  5  :  : 

8 

:  10 

6 

:  7  :  : 

12 

:  14 

Then 

2X4X6  : 

3X5X7 :: 

4X8X12 

Whence 

48 

105  : 

:  384  :  840. 

Let  a  :  b  '.  :  e  :  f, 
b  :  c  :  :  f  :  g, 
c  :  d  :  :  g  :  h, 


Then,  also,  a  :  d 


h. 


CI-        a     e     b     f 
Since  T=T>  -=-5  ^nd   , 
b     f    c      g  d 


DEMONSTRATION. 
C       g 


h 


,  we  obtain,  by  multiplying  the 


1  o       •  1        '^^c     efg         a     e 

alternate  tractions  tos-ether,  —-=-—-,  or  -^=t  >  thereiore,  a  :  a 
°  bed    jgh        d     k 


.  e  :  h. 

Let  2 
3 
4 


ILLUSTRATION   BY  NUMBERS. 


3 

4 

12 


6) 

8  (  Then  2  :  12  :  :  4  :  24. 


24 


By  multiplying  the  alternate  fractions,  we  have 

2X3X4  :  3X4X12  :  :  4x6x8  :  6x8x24. 
Whence,     24  :  144  :  :  192  :  1152,  or  2  :  12  :  :  4  :  24. 


RATIOS.  21o 

Prop.  XV.  If  there  are  any  number  of  quantities  more  than 
two,  and  as  many  others,  which,  taken  two  and  two,  in  cross 
order,  are  proportionals,  then  inversely,  by  equality,  are  the 
extreme  terms  in  the  first  set  proportional  to  the  extreme 
terms  in  the  second. 

Let  a,  h,  c,  d,  be  any  number  of  terms,  and  e,  jT,  g,  h,  as  many 
others,  and 

Let  a  :  h  :  :  g  '.  h  \ 

b  :  c  :  :  f  :  g  >  Then,  also,  a  :  d  :  :  e  :  k. 
c  :  d  :  :  e  :  f  ) 

DEMONSTRATION. 

Since -=^,  -=-,  and  -=-%,  by  multiplyino;   the  alternate 
b      k  c      g  d     f      "^  r  J    o 

fractions  together,  we  obtain 

ahc     gfe         a  e 

bed     hgf       d  It 

Therefore,  a  :  d  :  :  e  :  h. 

ILLUSTRATION  BY   NUMBERS. 


Let  2 


o 


3  :  4 

4  :  3 


Then,  2  :  3  :  :  8  :  12. 


2X3X4  :  3X4X3  :  :  8x6x8  :  12x8x6. 
Whence,  24  :  36  :  :  384  :  576. 

^j  dividing  the  first  two  terms  by  12,  and  the  last  two  by  48, 
we  obtain  2  :  3  :  :  8  :  12. 

Prop.  XYI.  When  four  quantities  are  proportionals,  if  the 
first  and  second  are  multiplied  or  divided  by  the  same  quantity, 
and  also  the  third  and  fourth  by  the  same  quantity,  the  resulting 
quantities  will  be  proportionals. 

Let  a  '.  b  '.  \  c  '.  d ;  then,  also,  ma  :  mb  :  :  nc  :  iid. 

DEMONSTRATION. 
CL        C 

Since  T=y  "^^  multiply  both  terms  of  the  first  by  m,  and 

both  terms  of  the  last  by  7i,  and  we  obtain  — -= — . ; 

mo     7id 


214  ALGEBRA. 

Therefore,  ma  :  mh  :  :  nc  :  7id, 

where  m  and  n  may  be  any  quantities,  either  integral  or  frac- 
tional. 

ILLUSTBATION   BY  NUMBERS. 

Let  2  :  4  :  :  3  :  6.  Now,  if  we  multiply  the  first  two  num- 
bers by  7,  and  the  last  two  numbers  by  9,  their  products  will  be 
proportionals.     Thus, 

2X7  :  4X7  :  :  3x9  :  6x9=14  :  28  :  :  27  :  54; 
and  if  any  other  numbers  were  taken  instead  of  7  and  9,  the 
products  would  be  proportionals. 

Prop.  XVII.  When  four  quantities  are  proportionals,  if  the 
first  and  third  are  multiplied  or  divided  by  the  same  quantity, 
and  also  the  second  and  fourth  by  the  same  quantity,  the  re- 
sulting quantities  will  be  proportionals. 

Let  a  :  b  :  :  c  :  d,  then,  also,  ma  :71b::  mc  :  nd. 

DEMONSTBATION. 

a     c  'tn 

Since  -7=-j-,  multiply  both  these  quantities   by  — ,  and  we 

obtain  — y-= — r,  therefore,  ma  :  Tib  :  :  mc  :  7id,  where  m  and 
no      nd 

n  may  be  any  quantities,  either  integral  or  fractional. 

ILLUSTRATION   BY   NUMBERS. 

Let  12  :  4  :  :  18  :  6,  and  we  will  multiply  the  first  and  third 
by  2,  and  the  second  and  fourth  terms  by  4.        "^ 

Thus,  12X2  :  4x4  :  :  18x2  :  6x4=24  :  16  :  :  36  :  24. 

It  is  evident  these  terms  are  proportionals  ; 

24     36        12     12 
^''  16=24'  ''  -S=-S- 

And  if  we  divide  the  first  and  third  terms  by  3,  and  the  second 
and  fourth  terms  by  2,  their  quotients  wiU  be  proportionals. 

Thus,  124-3  :  4-^2  :  :  18-^3  :  6-7-2. 

Or  4  :  2  :  :  6  :  3. 


RATIOS.  .  215 

4     6 
Whence,  o"^^* 

If  any  other  numbers  be  taken  for  multipljing  or  dividing, 
the  result  will  be  the  same. 

Prop.  XVIII.    If  four  quantities  are  proportionals,  the  like 
powers  or  roots  of  these  quantities  are  also  proportionals. 
heta:b::c:d;  then,  also,  a"'  :  h""  :  :  e"  :  d'". 
(Z      c 

Since  -=3,  raise  each  of  these  fractions  to  the  power  ex 
o     d 

t)  =  (  l )  '  "^^^  1^'^1~^'  therefore,  c"'  : 

IT-  :  :  c'"  :  d"^,  where  m  may  be  any  quantity,  either  integral  or 
fractional. 

ILLUSTRATION. 

Let  2  :  3  :  :  4  :  6,  then  2^  :  3'^  :  :  4^  :  61     If  we  raise 
each  of  these  terms  to  the  third  power,  the  result  will  be 
2X2X2=8  :  3x3x3=27  :  :  4x4x4=64  :  6x6x6=216. 

That  8,  27,  64,  and  216,  are  proportionals,  is  evident  from 

8       64 
the  fact  that  ^-=7tt:^,  and,  beino;  reduced  to  their  lowest  terms, 
27     210  ° 

27~27' 

Prop.  XIX.  Of  any  number  of  quantities  in  continued  pro- 
portion, the  first  has  to  the  third  the  duplicate  ratio,  to  the 
fourth  the  triplicate  ratio,  to  the  fifth  the  quadruplicate  ratio, 
&c.,  of  that  which  it  has  to  the  second,  or  of  that  which  the 
second  has  to  the  third,  &c. 

Let      a  '.  b  :  :  h  :  c  :  :  c  :  d  :  :  d  :  e  i  :  e  :  f  :  :  &c.  &c. 
Then    a  :  c  :  \  a'  :  h'\  or  in  the  duplicate  ratio  oi  a  '.  h. 
a  :  d  :  :  a'  :  h"\  or  in  the  triplicate  ratio  of  a  :  h. 
a  :  e  :  :  a^  :  Ij\  or  in  the  quadruplicate  ratio  of  a  :  b. 

DEMOXSTRATION. 

1st.  a  :  b  :  :  b  :  c,  OX,  by  Prop.  XVIII.,  a^  :  h-  :  :  h'  :  c" ; 
but,   by    Prop.    II.,   h'z=zac,    therefore,   a^   :    P    :  :    ac   :   c-. 


216  ALGEBRA. 

ov  a^  :  P  :  :  a  :  c,  hence  a  :  c  :  :  a^  :  b^ ;  also,  a^  :  ac  :  :  b^  :  c^ ; 

therefore,  a  :  c  :  :  b^  :  c^. 

2d.  a  :  c  :  :  a^  :  b-;  but  c  :  d  :  :  a  :  b  ;  therefore, 

a  :  d  :  :  a'  :  b^  :  :  P  :  c^  :  :  c'  :  (f. 

3d.  But  d  '.  e  '.  \  (T  \  W  \  therefore, 

a  :  e  :  :  ft^  :  ^-^  :  :  34  .  ^4  .  .  ^4  .  ^4  .  .  ^4  .  g4_ 

The  above  may  be  easily  illustrated  by  numbers. 

PROBLEMS   FOR   PROPORTION. 

1.  Divide  50  into  two  such  parts  that  the  greater,  increased 
by  3,  shall  be  to  the  less,  diminished  by  3,  as  3  to  2. 

Let  X  =  the  greater  number,  and  50 — x  the  less. 
Then  a:+3  :  50— a;— 3  :  :  3  :  2. 
Multiplying  extremes,     2x-\-Q=zlbQ — 3z — 9. 
Transposing,  5:r=135. 

Dividing,  a;=27,  the  greater. 

And  50-27=23,  the  less. 

2.  What  number  is  that  to  which  if  3,  8,  12,  and  20,  be 
severally  added,  their  sums  shall  be  j)roportional  ? 

Let  X  =  the  number. 

Then,  a;+3  :  x-j-8  :  :  x-{-12  :  2:+20. 

Multiplying  extremes,  a;2+23a:+60=2-+20:e+96. 
Transposing,  23z— 20^'=96— 60. 

Dividing,  2:=12.     A7is. 

VERIFICATION. 

12+3  :  12+8  :  :  12  +  12  :  12+20=15  :  20  :  :  24  :  32. 

3.  If  Mars,  when  in  opposition  to  the  sun,  is  49,000,000  miles 
from  the  earth,  and  the  quantity  of  matter  in  the  earth  is  11 
times  greater  than  that  in  Mars,  at  what  distance  from  the  earth, 
in  a  direction  towards  Mars,  will  a  body  remain  at  rest  ?  See 
Art  218. 

Let  X  =  the  distance  from  the  earth. 

Then  49,000,000— a;  =  the  distance  from  Mars. 

And  let  ^=49,000,000. 

Then,  a;2  :   {a—xY  :  :  1   :  11. 


RATIOS.  21' 

Multiplying  extremes,  llx'=a^ — 2ax-\-2^. 
Transposing,  10x^-{-2ax=a'^. 

2 

Reducing,  x--{-~z=—. 

Completing  the  squares,  x-+-^^+—=-+—=-^-^. 


a 


Evolving,  x-\-—=—Ayna\ 


a     .-^-^     a 


Transposing,  &c.  a:=— a/11 — — . 

And,  by  supplying  the  value  of  a,  we  have 


10^ 


('ll(49,000,000)A-iM^M5?=ii,351,430miles. 

Ans. 


4.  There  are  two  numbers  which  are  to  each  other  as  5  to  3  ; 
and,  if  4  be  added  to  the  greater  and  8  to  the  less,  they  will 
then  be  to  each  other  as  6  to  5.     What  are  the  numbers  ? 

Ans.  20  and  12. 

5.  Divide  the  number  GO  into  two  such  parts  that  their  pro- 
duct shall  be  to  the  difference  of  their  squares  as  2  to  3. 

A71S.  40  and  20. 

6.  I  have  two  square  house-lots,  which,  together,  contain  208 
square  rods;  and  the  area  of  the  greater  is  to  the  area  of  the 
less  as  9  to  4.  IIow  many  more  square  rods  are  there  in  the 
greater  than  in  the  less  ?  Ans.  80  square  rods. 

7.  The  product  of  two  numbers  is  12,  and  the  difference  of 
their  cubes  is  to  the  cube  of  their  difference  as  13  to  4.  \Vhat 
are  the  numbers  ?  A7is.  2  and  6. 

8.  Divide  the  number  100  into  two  such  parts  that  G  times 
their  product  shall  be  to  the  sum  of  their  squares  as  24  to  17. 
What  are  those  parts  ?  Ans.   80  and  20. 

9.  There  are  two  numbers,  whose  pAduct  is  35,  and  the  dif- 
ference of  their  squares  is  to  the  square  of  their  difference  as  G 
to  1.     What  are  the  numbers  ?  Aiis.  7  and  5. 

19 


218  ALGEBRA. 

10.  There  are  two  numbers  in  the  triplicate  ratio  of  4  to  1. 
whose  mean  proportional  is  32.     What  are  the  numbers  ? 

Ans.  25 G  and  4. 

11.  Divide  20  into  two  such  numbers,  that  the  quotient  ot 
the  greater  divided  by  the  less  shall  be  to  the  quotient  of  the 
less  divided  by  the  greater  as  9  to  4.    What  are  those  numbers  ? 

Am.  12  and  8. 

12.  Divide  26  into  three  such  parts,  that  the  first  shall  have 
the  same  ratio  to  the  second  that  the  second  has  to  the  third, 
and  that  the  first  term  shall  be  ^  the  third  term. 

A.ns.  2,  6,  and  18. 


SECTION  XX. 

ARITHMETICAL   PROGRESSIOIf. 

Art.  251.  An  Arithmetical  Progression  is  a  series  of  num- 
bers or  quantities,  increasing  or  decreasing  by  a  constant 
difierence. 

It  is  sometimes  called  Fj'ogression  by  Difference. 

252t  The  constant  difi'erence  is  called  the  Common  Difference^ 
or  ratio  of  the  progression. 

Ratio  here  used  is  an  Arithmetical  rate. 
Thus,  let  there  be  the  two  following  series. 

(1)   (2)   (3)  (4)   (5)  (G)   (7)   (8) 
First  series,         1,    4,    7,  10,  13,  16,  19,  22=92. 

Second  series,   30,  26,  22,  18,  14,  10,    G,    2=128. 

253.  The  numbers  which  form  the  series  are  called  the  terni^ 
of  the  progression. 

254.  The  first  is  ci^led  an  ascending  series  of  progression, 
where  the  first  term  is  1,  the  common  diiference  3,  the  number 
of  terms  8,  the  last  term  22,  nnd  the  snui  of  tlie  series  92. 


ARITHMETICAL     PROGRESSION.  219 

255.  The  second  is  called  a  descending  5eWes  of  progression, 
where  the  first  term  is  30,  the  common  difi'erence  — 4,  the 
number  of  terms  8,  the  last  term  2,  and  the  sum  of  the  series 

128. 

256.  The  first  and  last  terms  of  the  progression  are  called 
extremes^  and  the  other  terms  are  the  means. 

257.  The  number  of  common  difi"erences  in  any  number  of 
terms  is  one  less  than  the  number  of  terms. 

Hence,  if  there  be  8  terms,  the  number  of  common  differences 
will  be  7,  and  the  sum  of  the  differences  will  be  equal  to  the 
difference  of  the  extremes. 

We  therefore  infer,  that  if  the  difference  of  the  extremes  be 
added  to  the  first  term,  the  sum  will  be  the  last  term ;  also,  if 
the  difference  of  the  extremes  be  taken  from  the  last  term,  the 
remainder  will  be  the  first  term. 

258.  Also,  if  the  sum  of  the  common  differences  be  divided 
by  the  number  of  common  differences,  the  quotient  will  be  the 
common  difference. 

To  illustrate  this,  we  will  examine  the  following  series  : 

(1)     (2)     (3)     (4)     (5)      (6)      (7) 
2,      5,      8,     11,    14,    17,     20. 

Here  the  first  term  is  2,  the  last  term  20,  the  number  of 
terms  7,  and  the  common  difference  3. 

Now,  if  we  had  only  the  first  term,  number  of  terms,  and 
common  difference,  to  find  the  last  term,  we  should  have  only  to 
add  the  difference  of  the  extremes  to  the  first  term. 

The  common  difference  is  3 ;  and,  as  there  are  7  terms,  the 
number  of  common  differences  is  6.  The  difference  of  the 
extremes  will,  therefore,  be  6X^=18,  and  the  last  term  will  be 
2+18=20. 

Hence,  having  the  first  term,  common  difference,  and  number 
of  terms  given,  to  find  the  last  term,  we  have  the  following 

Rule.  Multiply  the  number  of  terms,  less  one,  by  the  common 
difference,  and  to  the  product  add  the  first  term. 


220  ALGEBRA. 

Again,  if  we  invert  the  terms,  we  have 

(1)       (2)        (3)       (4)       (5)     (6)     (7) 
20,      17,      14,      11,       8,      5,      2. 

Here  we  have  20  for  the  first  term,  — 3  for  the  common  dif- 
ference, and  7  for  the  number  of  terms,  to  find  the  last  term. 
6X— 3=— 18  ;  20—18=2  the  last  term. 

The  pupil  will  perceive  that  18  is  a  negative  term;  and  to 
add  a  negative  term  to  a  positive  is  to  write  their  difference. 

Again,  we  have  given  the  extremes  2  and  20,  and  number  of 
terms  7,  to  find  the  common  difference. 

Here  the  number  of  common  differences  is  6 ;  for  we  have 
before  shown  that  the  number  of  common  differences  is  always 
one  less  than  the  number  of  terms;  therefore,  18-7-6=3,  the 
common  difference. 

259»  The  principles  of  an  arithmetical  progression  may  be 
well  illustrated  by  literal  terms. 

Let  a  be  the  first  term  of  an  ascending  series,  and  d  the 
common  difference ;  then  the  second  term  will  be  a-\-d,  and  the 
the  third  term  a-\-2d,  and  the  series  will  be 

(1)  (2)  (3)  (4)  (5)  (6) 

a,         a-{-d,         a-\-2d,         a-\-od,         a-{-4:d,         a-\-bd. 

If  it  be  required  to  form  a  descending  series,  when  the  first 
term  is  a  and  the  common  difference  — d,  it  will  be  thus  : 
(1)  (2)  (3)  (4)  (5)  (6) 

a,         a — d,         a — Id,         a — 3^,         a — 4c?,         a — bd. 

260t  It  is  evident  that  the  last  term  in  both  series  is  equal 
to  the  first  term  with  the  common  difference  repeated  as  many 
times,  wanting  one,  as  there  are  terms  in  the  series. 

Hence,  if  n  represent  the  number  of  terms,  the  following  will 
be  the  formula  to  find  X,  the  last  term. 

h=^a-\-{n — V)d. 

EXAMPLES. 

1.  If  the  first  term  be  7,  the  common  difference  4,  and  the 
number  of  terms  20,  required  the  last  term. 

i=a+(7i-]y=7+(20-l)4=83.     Am. 


ARITHMETICAL    PROGRESSION.  221 

2.  If  the  first  term  is  3,  the  common  difi"erence  5,  required 
the  50  th  term. 

jL=a+(w-l)6^=34-(50— 1)5=248.     Ans. 

3.  If  the  first  term  is  90,  the  common  difi"erence   —7,  re- 
quired the  10th  term. 

X=a+(7i-l)(-^)=90+(10-l)(-7)=27.     Ans. 

4.  If  the  first  term  is  f ,  the  common  dijQFerence  1^,  what  is 
the  20th  term  ? 

i=fl+(7z-l)^=f+(20-l)l^=26TV.     Ans. 

5.  If  the  first  term  is  18,  the  common  difi"erence  —4,  what  is 
the  10th  term? 

X=a+(?i-l)(— ^)=18-f(10— 1)(-4)=-18.     Ans. 

261 .  The  formula  for  obtaining  the  first  term,  a,  is  obtained 
from  the  former  by  transposition. 

Thus,  if  X=a+(?z — l)d,  then,  by  transposition, 

a=L — (n — l)d. 

6.  If  the  last  term  is  25,  the  number  of  terms  6,  and  the 
common  difierence  2,  required  the  first  term. 

fl=i-(7i—l)^=25-(G— 1)2=15.     Alls. 

7.  If  the  last  term  is  50,  the  common  difierence  6,  the  number 
of  terms  10,  required  the  first  term. 

a=L—{7i—'[)d=^0—(iO-l)Q=-4.     Ans. 

8.  If  the  last  term  is  27,  the  common  diff'crence  2^,  number 
of  terms  10,  required  the  first  term. 

a=L— {7i-l)d=27 -{10— 1)2^=U.     A?is. 

262.  The  formula  for  obtaining  the  common  difi"ercnce,  d,  is 
obtained  from  the  first  by  transposition  and  division. 

Thus,  L=a^{?i—l)d. 

Then,  by  transposition,         L — a={n — \)d. 

And  by  division,  —=d, 

•^  71  —  1 

L — a, 
Changing  terms,  d= r- . 


222  ALGEBRA. 

9.  If  the  extremes  are  6  and  30,  and  the  number  of  terms 
13,  what  is  the  common  difference  ? 

,     L — a     30 — 6     ^        . 
n — 1      Id — 1 

10.  If  the  extremes  are  ^  and  15j,  and  the  number  of  terms 
11,  what  is  the  common  difference  ? 

,     L—a     15|— f     ^ 

d= -=-^--i=li      Am. 

n — 1       11  —  1 

263.  The  formula  for  obtaining  the  number  of  terms  may  be 
obtained  from  the  first  formula. 

Thus,  L=a-Y{n—V)d. 

By  transposition,      L — a={n — V)d. 

By  division,  — ; — =?2 — 1. 

-^  d 

By  transposition,     — - — \-\=.7i, 

T 

Changing  terms,       n= — j-1. 

11.  If  the  extremes  m'e  3  and  39,  and  the  common  difference 
2,  what  is  the  number  of  terms  ? 

L-a  .  .      39—3 


n=—, [-1= — \ f-l=19-     ^-^is. 

rl  V. 


12.  If  the  first  term  is  5,  the  last  term  89,  the  common  dif- 
ference 7,  required  the  number  of  terms. 

L — a  ,  ^      89 — 5     ^      TO       . 
n=— — ^^="7 [-1=13.     Alls. 

Having,  therefore,  any  three  of  the  four  terms  given,  the 
other  may  be  found,  as  we  have  demonstrated  above,  by  the 
following 

yORMULiE. 

(1.)  To  find  the  last  term. 

L=a-j-{n — l)d. 


ARITH.AIET1CAL     PROGRESSION.  223 

(2.)  To  find  the  first  term. 

a=L — [n — l)d. 

(3.)  To  find  the  common  difference. 

J     L — a 

d= =. 

n — 1 

(4.)  To  find  the  number  of  terms. 

L — a 


■1. 


d 

When  the  series  are  descending,  the  unknown  difi"erence  is  a 
minus  quantity  in  the  1st  and  2d  formula) ;  thus,  — d. 

13.  A  man  travelled  10  days;  the  first  day  he  went  8  miles, 
the  second  day  13  miles,  and  thus  increased  his  distance  each 
day  5  miles.     How  far  did  he  travel  the  last  day  ? 

Ans.  53  miles. 

14.  John  Smith's  family  expenses  for  the  first  year  were 
$500 ;  but,  after  he  had  been  married  12  years,  he  found  his 
last  year's  expenses  to  have  been  $1325.  By  how  much  did  he 
increase  his  expenses  yearly  ?  A?is.  $75. 

15.  A  man  set  out  from  Boston  to  travel  into  the  country ; 
the  first  day  he  travelled  12  miles,  the  second  day  9  miles,  the 
third  day  6  miles,  and  thus  continued  to  travel  each  day  3 
miles  less  than  the  preceding.  How  far  did  he  go  the  tenth 
day  ?  Ans.   — 15  miles. 

261.  To  find  the  sum  of  the  series. 

ARITHMETICAL   SERIES. 

(1)  (2)  (3)  (4)   (5)  (G) 
Let  2,    5,    8,  11,  14,  17,  be  the  series. 

And  17,  14,  11,    8,    5,    2,  same  series  inverted. 

19,  19,  19,  19,  19,  19,  sum  of  both  series. 

LITERAL    SERIES. 

(1)  (2)  (3)  (1)  (5)  (G) 

Let   a,  a-{-d,     a-{-2d,     a-\-od,     a-{-4:d,     a -{-5c?  bo  a  series. 

And  a-\-^d,     a-{-4d,  a-{-Sd,     a-\-2d,     a-{-d,       a 


same  series 
inverted. 


2a-\-bd,  2a-[-bd,  2a-\-bd,  2a-[-bd,  2a-\-bd,  2a-f  5^,  sum  of 
both  series. 


224  ALGEBRA. 

We  perceive,  from  the  above  aritlimetical  and  literal  series, 
that  the  sum  of  the  extremes  is  equal  to  the  sum  of  any  two 
of  the  means  equally  distant  from  each  extreme ;  and  that,  by 
adding  the  two  series  in  their  present  arrangement,  we  have  the 
same  number  for  the  same  successive  terms ;  also,  that  the  sum  of 
both  series  is  twice  the  sum  of  either  series.  Therefore,  if  19, 
the  sum  of  the  extremes  in  the  arithmetical  series,  be  multiplied 
by  0,  the  number  of  terms,  the  product  will  be  the  sum  of  both 
series.  Thus,  19x^=114,  sum  of  both  series.  Therefore, 
114-f-2=57  will  be  the  sum  of  either  series. 

Again,  2a-]-^d  is  the  sum  of  the  extremes  in  the  literal 
series;  and,  if  this  sum  be  multiplied  by  6,  the  number  of 
terms,  the  product  will  be  the  sum  of  both  series.  Thus, 
{2a-{-M)Q=12a-{-S0d,  sum  of  both  series.  And  (12a+30*i) 
-^2=Qa-{-l^d,  the  sum  of  either  series. 

Therefore,  in  all  cases,  we  find  that  the  simi  of  the  series  is 
equal  to  the  sum  of  the  extremes  multiplied  by  half  the  number 
of  terms ;  or,  the  number  of  terms  multiplied  by  half  the  sum 
of  the  extremes. 

If,  therefore,  the  sum  of  any  series  be  denoted  by  S,  the  first 
term  by  a,  the  last  term  by  L,  and  the  number  of  terms  by  w, 
the  following  will  be  the  formula  for  obtaining  its  value  : 

Therefore,  if  the  extremes  and  the  number  of  terms  are  given 
to  find  the  sum  of  the  series,  we  adopt  the  following 

EuLE.  Multiply  /lalf  the  sum  of  the  extremes  hij  the  number 
of  terms. 

The  two  following  formulae,  or  equations,  contain  five  quan- 
tities :  a,  the  first  term  of  a  progression ;  i,  the  last  term ;  d, 
the  common  difierence ;  n^  the  number  of  terms ;  and  »S,  the 
sum  of  the  series. 

If  any  three  of  these  be  given,  the  other  two  may  be  ob- 
tained. 

(1.)     L=a+{n-\)d.  (2.)     S=(^^^». 


S^f^^' 


AKITHMKTICAL    PKOGEESSION.  225 

265 1  The  pupil  will  find  that  twenty  different  cases  may  arise 
which  may  be  solved  by  difierent  combinations  of  the  above 
equations. 

To  find  n  in  the  last  equation. 


2 

By  multiplication,        ^Sz={^L-\-a)n. 

By  division,  -^^-- — =7^. 

L-\-a 

2S 
Therefore,  7i-. 


'L-{-a 

If,  therefore,  the  extremes  and  the  sum  of  the  scries  are  given 
to  find  the  number  of  terms,  we  divide  twice  the  sum  of  the 
series  by  the  sum  of  the  extremes. 

16.  Let  the  extremes  be  3  and  39,  and  the  sum  of  the  series 
399,  to  find  the  number  of  terms. 

2S       2x399     ^-        . 
L-\-a      o\}-\-6 

266.  To  find  the  last  term,  L,  from  the  second  equation. 

By  multiplication,  2S={L-\-a)n, 

By  division,  — =L-{-a. 

2S 

By  transposition,  a=L. 

n 

2S 
By  transposition  of  terms,        L=- a. 

Therefore,  having  the  first  term,  number  of  terms,  and  sum 
of  the  series,  given  to  find  the  last  term,  we  divide  twice  the 
sum  of  the  series  by  the  number  of  terms,  and  subtract  the  first 
term  from  the  quotient. 

267 »  To  find  the  first  term,  a,  from  the  second  equation. 


226  ALGEBRA. 

Multiplying,  2S={Li-a)n. 

2S 
Dividing,  — =L4-a. 

n 

Transposing,  L=a. 

Chandnsf  terms,  '  <2= L. 

°  n 

Therefore,  having  the  last  term,  number  of  terms,  and  sum  of 
the  series,  given  to  find  the  first  term,  we  divide  twice  the  sum 
of  the  series  by  the  number  of  terms,  and  subtract  the  last  term 
from  the  quotient. 

17.  Let  the  last  term  be  39,  number  of  terms  19,  and  the 
sum  of  the  series  399,  to  find  the  first  term. 

2S  2X399     OQ     o      4 

a= X(= — o9=D.     Atis. 

n  19 

268 1  To  find  the  common  diflerence,  d^  from  the  1st  and  2d 
equation. 

We  find  the  value  of  X,  in  the  first  equation,  to  be 
h=.(i-\-{n — \)d. 

Substituting  this  value  of  L  for  iS  in  the  2d  equation,  and 

then  transposing,  we  have 

^     2S—2an 
o= . 

?l(7l — 1) 

18.  If  the  first  term  is  5,  the  number  of  terms  15,  and  the 
sum  of  the  series  285,  what  is  the  common  difference  ? 

Ans.  2. 

19.  If  the  first  term  is  3,  the  number  of  terms  19,  and  the 
sum  of  the  series  399,  what  is  the  common  difference  ? 

Alls.  2. 

20.  If  the  first  term  is  7,  the  number  of  terms  8,  and  the 
sum  of  the  series  100,  what  is  the  common  difference  ? 

A?is.  li. 

Problems. 

1,  The  first  term  is  5,  the  common  difference  3.  What  is  the 
7th  term  ?  A?is.  23. 


ARITHMETICAL     PROGRESSION.  227 

2.  The  first  term  is  3,  the  common  difference  4^.  "What  is 
the  5th  term  ?  Ans.  20^. 

3.  The  first  term  is  18,  the  common  difference  ^.  What  is 
the  7th  term  ?  Am.  19J . 

4.  The  first  term  is  7,  the  common  difference  2^,  and  the 
number  of  terms  5.     Required  the  last  term.  Ans.  17. 

5.  The  first  term  is  ■'^,  the  common  difference  |.  What  is  the 
10th  term?  Am.  7^^-. 

6.  The  first  term  is  0,  the  common  difference  Ij-.  What  is 
the  20th  term?  Am.  28^. 

7.  The  first  term  is  10,  the  common  difference  — 2.  What  is 
the  4th  term  ?  Am.  4. 

8.  The  first  term  is  — 8,  the  common  difference  — 3.  What 
is  the  10th  term  ?  Am.  —35. 

9.  The  first  term  of  a  descending  series  is  85,  common  dif- 
ference 7.     Required  the  10th  term.  Ans.  22. 

10.  The  first  term  is  3i,  the  common  difference  2|-.  What  is 
the  5th  term,  and  the  sum  of  the  series  ?    Atis.  124-,  and  39^. 

11.  The  first  term  in  a  descending  series  is  2^,  the  common 
difference  is  ^.  What  is  the  10th  term,  and  the  sum  of  the 
series?  An^.  {,  and  13|. 

12.  The  first  term  is  a,  the  common  difference  is  d.  AVhat  is 
the  Tzth  term  ?  A7is.  a-\-d{?i — 1). 

13.  What  is  the  sum  of  the  odd  numbers  from  1  to  100  ? 

Am.  2500. 

14.  If  the  first  term  is  4^-,  the  common  difference  3^,  and 
number  of  terms  8,  what  is  the  sum  of  the  series  ?    Aiis.  134. 

15.  If  the  first  term  is  7,  the  common  difference  — 4,  and  the 
number  of  terms  6,  what  is  the  sum  of  the  series  ? 

A?is.  —18. 

16.  If  the  first  term  is  5,  the  last  term  19,  and  the  number 
of  terms' 6,  what  are  the  other  terms  of  the  progression? 

Am.  7|,  lOf,  13|,  IQi. 


228 


ALGEBRA . 


17.  If  the  extremes  are  —9  and  18,  and  the  number  of  terms 
5,  what  are  the  other  terms  of  the  progression  ? 

Atls    2i   4i   llJ- 

18.  If  the  last  term  of  an  ascending  series  is  20,  the  com- 
mon difference  5,  and  the  number  of  terms  8,  what  is  the  sum 
of  the  series  ?  Ans.  20. 

19.  There  is  a  number  consisting  of  three  digits  in  arith- 
metical progression,  whose  sum  is  12;  and,  if  896  be  added  to 
the  number,  the  digits  will  be  inverted.     What  is  the  number  < 

Am.  246. 

20.  There  is  a  certain  island  50  miles  in  circumference.  Two 
men,  A  and  B,  set  out  to  travel  round  it.  A  goes  10  miles 
each  day.  B  goes  2  miles  the  first  day,  5  miles  the  second  day, 
and  8  miles  the  third  day,  travelling  each  day  3  miles  further 
than  the  day  preceding.  How  far  will  A  and  B  be  apart  the 
8th  day  ?  Ans.  30  miles. 

21.  John  Smith  and  John  Jones  set  out  from  Boston  for  the 
city  of  Washington,  the  distance  being  440  miles.  Smith 
started  5  days  before  Jones,  and  travels  15  miles  per  day. 
Jones  travels  25  miles  the  first  day,  23  miles  the  second  day, 
and  21  miles  the  third  day,  travelling  each  day  2  miles  less 
than  the  preceding.  How  far  apart  will  Smith  be  from  Jones  at 
the  end  of  the  20th  day,  and  how  far  will  each  be  from 
Washington  ? 

Ans.  135  miles  apart.  Smith  140  miles  from  Washington. 
Jones  275  miles  from  Washington. 

22.  If  the  first  term  is  ^,  the  common  difference  — -^-,  and 
the  number  of  terms  20,  what  are  the  last  term  and  the  sum  of 
the  series?  .       (  Last  term,  — 2f. 

(  Sum  of  the  series,  — 21§. 

23.  If  one  extreme  is  ■^,  the  common  difference  — yL,  and 
the  sum  of  the  series  — 1^,  what  is  the  number  of  terms  ? 

Ans.  12. 

24.  If  the  first  term  is  -/g^,  last  term  2J-,  and  the  sum  of  the 
series  37,  what  is  the  number  of  terms  ?  Ans.  24. 


ARITHMETICAL     PROGRESSION.  229 

25.  If  the  first  term  is  3,  the  last  term  17,  and  the  number 
of  terms  29,  what  arc  the  terms  of  the  series  ? 

Am'.  3,  3^,  4,  U,  5,  5i,  &c. 

26.  The  sum  of  the  series  is  16^,  the  number  of  terms  10, 
and  the  common  difference  -],  to  find  the  first  term.      Ans.  4. 

27.  The  first  term  of  an  arithmetical  series  is  — 5,  the  com- 
mon difierence  1^;  what  is  the  9th  term  ?  Ans.  7. 

28.  What  are  the  three  means  between  — 1  and  15  ? 

A71S.  3,  7,  and  11. 

29.  The  first  term  is  1^,  number  of  terms  10,  and  the  sum  of 
the  series  6|.     What  is  the  common  difference  ?       A?is.  — ^. 

30.  There  are  three  numbers  in  arithmetical  progression 
(vhose  sum  is  10,  and  the  product  of  the  second  and  third  is 
33^.     What  are  those  numbers  ?  Atis.  — 3^,  3^,  and  10. 

31.  The  number  of  terms  of  an  arithmetical  progression  is 
equal  to  J-  the  common  difTerence,  the  last  term  is  equal  to  4 
times  the  first,  and  the  sum  of  the  series  is  equal  to  f  the 
square  of  the  first  term.  What  are  the  scries,  and  the  sum  of 
the  series  ? 

,  The  series,  20,  32,  44,  56,  68,  80. 
Atis. 


■  i 


Sum  of  the  series,  300. 


32.  There  arc  four  numbers  in  arithmetical  progression  whose 
sum  is  28,  and  the  sum  of  whose  squares  is  216.  What  are 
those  numbers  ?  Ans.  4,  6,  8,  and  10. 

33.  Find  three  numbers  in  arithmetical  progression  whose 
sum  is  9,  and  the  sum  of  whose  cubes  is  99. 

Ans.  2,  3,  and  4. 

34.  What  are  those  four  numbers  in  arithmetical  progression 
the  sum  of  the  sc[uares  of  whose  first  two  terms  is  34,  and  the 
sum  of  the  squares  of  the  last  two  is  130  ? 

An£.  3,  5,  7,  and  9. 

35.  A  certain  number  consists  of  three  digits,  which  are  in 
arithmetical  progression  ;  and,  if  the  number  be  divided  by  the 
sum  of  its  digits,  the  quotient  will  be  274,  but,  if  396  be  added 

20 


230  ALGEBRA. 

to  the  number,  the  digits  will  be  inverted.     Required  the  num- 
ber. Ans.  579. 

36.  What  are  those  four  numbers  in  arithmetical  progression 
the  sum  of  the  squares  of  whose  extremes  is  90,  and  the  sum  of 
the  squares  of  the  means  is  74  ?  Ans.  3,  5,  7,  and  9. 

37.  What  are  those  four  numbers  in  arithmetical  progression 
whose  sum  is  14,  and  whose  continued  product  is  120  ? 

Atis.  2,  3,  4,  and  5. 

38.  There  are  four  numbers  in  arithmetical  progression,  the 
product  of  whose  extremes  is  112,  and  that  of  the  means  120. 
What  are  the  numbers  ?  Ans.  8,  10,  12,  and  14. 

39.  A  and  B,  165  miles  from  each  other,  set  out  with  a 
design  to  meet.  A  travels  one  mile  the  first  day,  two  the 
second,  three  the  third,  and  so  on.  B  travels  20  miles  the  first 
day,  18  the  second,  16  the  third,  and  so  on.  How  soon  will 
they  meet  ?  Atis.  10  days,  or  33  days. 

40.  There  are  four  numbers  in  arithmetical  progression,  whose 
continued  product  is  1680,  and  common  difference  is  4.  Re- 
quired the  numbers.  Ans.  14,  10,  6,  2. 

41.  Five  persons  undertake  to  reap  a  field  of  87  acres.  The 
five  terms  of  an  arithmetical  progression,  whose  sum  is  20,  will 
express  the  times  in  which  they  can  severally  reap  an  acre,  and 
they  all  together  can  finish  the  job  in  60  days.  In  how  many 
days  can  each,  separately,  reap  an  acre  ? 

A71S.  2,  3,  4,  5,  6  days. 

42.  A  gentleman  set  out  from  Boston  for  New  York.  He 
travelled  25  miles  the  first  day,  20  miles  the  second  day,  each 
day  travelling  5  miles  less  than  the  preceding.  How  far  was 
he  from  Boston  at  the  end  of  the  eleventh  day  ?  A?is. 

43.  Suppose  a  number  of  stones  were  laid  a  rod  distant  from 
each  other  for  twenty  miles,  and  the  first  stone  a  rod  from  a 
basket.  What  length  of  ground  will  that  man  travel  over,  who 
gathers  them  up  singly,  returning  with  them,  one  by  one,  to  the 
basket?  Atis.  128,060  miles,  2  rods. 


ARITHMETICAL     PROGRESSION. 


231 


There  are  twenty  diflferent  eases  in  Arithmetical  Progression, 
all  of  which  are  exhibited  in  the  followinc^  Table. 


No. 


Given. 

a,  d,  n 
a,  d,  S 

a,  71,  S 
d,  71,  S 


a,  d,  n 

a,  d,  I 

a,  I,  n 

d,  n,  I 


Requir'd. 


Formulae. 


I  =a-\-{n — \)d. 
I  I  =^-JLd±^'ldS+{a-^df. 

n 


n 


2 


S=^?^[2a+(7^— 1)^]. 


S=- 


1d   ' 


S=ln{^1l—{ii—\)d). 


9     I   a,  ?z,  I 


10 


12 


18 
14 
15 
IG 


17 

18 
19 
20 


a,  n,  o 


n    !  «,  z,  s 


n,  I,   S 


I — a 
n — 1' 


d: 


d  = 


2S-2an 
n{7i — 1) ' 


2S—l-a 
27il-2S 


d= 


n[n — 1)* 


d,  71,  I 

d,  71,  S 

d,  I,  S 

71,  I,  S 


a,  d,  I 

a,  d,  S 

a,  I,  S 

d,  I,  S 


a=l—{ii—\)d. 
S      [ii—\)d 


71 


a=.id±^  {l-]-U)-—2dS. 


2S 
a= L 

71 


71 


I— a 


7l  =  ± 


V  ('2a—dY-ir^dS—2a-\-d 


2d 


2S 


71= 


2lJ^d±/s/ {2l-j-d)'—MS 
2d  ■ 


232  ALGEBRA. 


SECTION   XXI. 

GEOMETRICAL   PROGRESSION,    OR   PROGRESSION   BY 
QUOTIENT. 

Art.  269t  When  there  are  three  or  more  numbers,  such  that 
the  same  quotient  is  obtained  by  dividing  the  second  by  the 
first,  and  the  third  by  the  second,  and  the  fourth  by  the  third, 
&c. ;  or,  such  that  they  increase  or  decrease  by  a  constant 
multiplier,  they  are  said  to  be  in  Geometrical  Progression,  and 
are  called  a  Geometrical  Series.  Thus, 
(1)     (2)  (3)  (4)    (5)    (6) 

(1.)        2,      6,  18,  54,  162,  486  =  728,  sum  of  the  series. 

(2.)    486,  162,  54,  18,      6,      2  =  728,  sum  of  the  series. 

The  first  is  called  an  ascending  series,  and  the  second  a  de 
scending  series. 

In  the  first  the  quotient  or  multiplier  is  3,  and  it  is  called 
the  ratio.     In  the  second  the  ratio  is  ^. 


The  first  and  last  terms  of  a  series  are  called  the  ex- 
tremes, and  the  others  are  the  means. 

271.  It  will  readily  be  perceived,  in  either  of  the  above  series 
that  the  product  of  the  extremes  is  equal  to  the  product  of 
any  two  of  the  means  equally  distant  from  the  extremes.  Thus, 
2x486=6x162=18x54=972. 

272a  If  there  are  only  three  terms,  the  product  of  the  ex- 
tremes is  equal  to  the  square  of  the  second  term. 

273.  It  is  evident,  by  examining  either  the  above  series,  that 
any  term  may  be  obtained  by  multiplying  the  first  term  by  the 
ratio  as  many  times,  wanting  one,  as  there  are  terms  required. 

If,  therefore,  the  1st  term  is  2,  and  the  ratio  3,  and  we  wish 
to  obtain  the  6th  term,  we  have  only  to  multiply  the  1st  term, 
2,  by  the  ratio  3,  five  times. 

Thus,         2X3X3X3X3X3=486,  the  6th  term. 


GEOMETRICAL     PROGRESSION.  233 

The  above  may  be  generalized  in  the  following  manner : 

Let  a  =  first  term  of  a  series. 

L  =  the  last  term. 

r  =  the  ratio. 

n  =  the  number  of  terms. 

S  =  the  sum  of  the  series. 

(1)  (2)    (3)     (4)      (5)    (G) 
Then    a,   ar,   ar^,   ar',   ar^,   ar^,    &c.,   may    represent    any 
geometrical  series ;  and,  if  r,  the  ratio,  is  considered  as  more 
than  a  unit,  the  series  is  ascendiiig  ;  but,  if  r  is  less  than  a  unit, 
the  series  is  descending. 

The  exponent  of  r  in  the  second  term  is  1,  in  the  third  term 
2,  in  the  fourth  term  3,  in  the  fifth  term  4,  and  so  on  ;  there- 
fore, the  exponent  of  r  in  the  last  term  will  always  be  one  less 
than  the  number  of  terms.  The  exponent  of  the  nih.  term  in  the 
above  series  would  therefore  be  Gir""^ 

274.  If,  therefore,  in  any  series  the  number  of* terms  be 
denoted  by  ti,  and  the  last  term  by  L,  the  following  will  be  the 
formula  for  finc^ng  the  last  term  : 

(1.)  L=ar'-\ 

And     L=r"~^j  when  the  first  term  is  a  unit. 

In  the  above  equation  we  have  four  quantities,  a,  L,  r,  and  ?i  ; 
<^nd,  if  any  three  of  them  be  given,  the  others  may  be  obtained 
ds  follows : 

To  find  a,  the  first  term,  we  divide  both  terms  of  the  above 
equation  by  7-"~^,  and  transpose  the  terms  ;  and  we  have 

(2.)  «=^- 

To  obtain  r,  the  ratio,  we  divide  the  terms  of  the  1st  equa- 
tion by  a,  extract  the  (7i— l)th  root,  and  transpose  the  terms ; 
and  we  have 

To  find  71,  we  shall  show  when  we  come  to  treat  of  exponential 
quantities. 

20^ 


234  ALGEBRA. 

EXAMPLES. 

1.  If  the  first  term  is  7,  the  ratio  3,  and  the  number  of  terms 
5,  required  the  last  term. 

i=«r"-i=7(3)4=:567.     Am. 

2.  If  the  first  term  is  1,  the  ratio  5,  and  the  number  of  terms 
5,  what  is  the  last  term  ? 

i:=r"-^=54=625.     A71S. 

3.  If  the  last  term  is    405,  the  ratio  3,  and  the  number  of 
terms  5,  what  is  the  first  term  ? 

L       405     ,       , 

4.  If  the  last  term  is  8,  ratio  5,  and  the  number  of  terms  4, 
what  is  the  first  term  ? 

Z         8         8         ^ 

^.n-l         54-1         125 

5.  If  the  first  term  is  5,  the  last  term  1215,  and  the  number 
of  terms  6,  what  is  the  ratio  ? 

/L\-L-     /1215\J_     ^,  i     ^       , 
r={  -  l"-i=  — ^—  V'-i  =243^=3.     A71S. 


i!)^-iT} 


27 
6.  If  the  first  term  is  ^,  the  last  term  — — ,  and  the  number  of 

o2U 

terms  4,  what  is  the  ratio  ? 


r= 


7.  If  the  first  term  is  y^g,  the  last  term  64,  and  the  number 
of  terms  6,  required  the  ratio.  Ans.  4. 

8.  If  the  last  term  is  135,  the  number  of  terms  4,  the  ratio 
3,  what  is  the  first  term  ?  A71S.  5. 

275.  To  find  any  number  of  geometrical  means  between  any 
two  given  numbers. 

n-l       IJ^ 

In  the  3d  formula,  we  found  r=         — . 


GEOMETRICAL     PllOGRESSION.  235 

If  we  let  m  represent  the  number  of  means,  then  ?n-\-2=i7i , 
for  the  number  of  terms  is  always  two  more  than  the  number 
of  means. 


Therefore,  (  — \'-J= (^  -  \  "'+i. 

Consequently,  ?•=  f  —  j"'+^ 


276.  Having,  therefore,  the  extremes  given  to  find  any  num- 
ber of  means,  we  divide  the  greater  extreme  or  number  by  the 
less  extreme,  and  extract  that  root  of  the  quotient  denoted  by 
the  number  of  means  plus  1.     This  root  is   the  ratio ;    and 
having  the  ratio,  the  means  are  readily  obtained. 

EXAMPLES. 

9.  Find  two  geometrical  means  between  6  and  162. 
lG2-r-6=27  :  /^27=3,  the  ratio  ;  6x3=18,  the  first  mean  ; 

18x3^54,  the  second  mean. 

10.  What  is  the  geometrical  mean  between  18  and  882  ? 
882-^18=49  :  V49=7,  the  ratio;    18x7=126,  the  geo- 
metrical mean. 

11.  Required  the  five  geometrical  means  between  1  and  64. 

Am.  2,  4,  8,  16,  32. 

12.  A  has  a  piece  of  land,  which  is  18  rods  wide,  and  288 
rods  long.  Required  the  side  of  a  square  piece  that  shall  con- 
tain an  equal  number  of  square  rods.  Ans.  72  rods. 

277.  To  find  the  sum  of  all  the  terms  of  a  geometrical  series. 
Let  the  following  be  the  series : 

(1.)  2,  6,  18,  54,  162. 

By  examining  this  series,  we  find  the  first  term  2,  the  ratio  3, 
and  the  last  term  162. 

If  we  multiply  each  term  in  the  series  by  the  ratio  3,  we 
obtain 

(2.)  6,  18,  54,  162,  486. 

It  is  evident  that  the  simi  of  this  last  series  is  three  times  the 


236  ALGEBRA. 

former ;  therefore  the  difference  between  them  will  be  equal  to 
twice  the  sum  of  the  first  series.     Thus, 

From         6,  18,  54,  162,  486,  second  series. 

Take     2,  6,  18,  54,  162,  first  series. 

— 2  486=484,  difference  of  the  series. 

From  the  above  operation,  it  appears  that  484  is  twice  the 
sum  of  the  first  series ;  and,  therefore,  484-7-2=242  is  the 
sum  required. 

Bj  examining  the  process,  we  perceive  that  242  is  obtained 
by  multiplying  the  last  term  of  the  first  series,  162,  by  the 
ratio  3,  and  subtracting  from  the  product  the  first  term  2,  and 
dividing  the  remainder,  484,  by  2  a  number  which  is  one  less 
than  the  ratio.     Hence  the  propriety  of  the  following 

KuLE.  Multiply  the  last  term  by  the  ratio,  find  the  difference 
between  this  product  and  the  first  term,  divide  this  remainder  by 
the  difference  between  the  ratio  and  unity,  and  we  have  the  sum 
of  the  series. 

278.  We  may  generalize  the  above,  as  follows : 

Let  a  represent  the  first  term  of  a  geometrical  series,  r  the 

ratio,  L  the  last  term,  n  the  number  of  terms,  and  S  the  sum  of 

the  series.     Then 

(1.)  S=a-\-ar-\-ar^-\-ar^-\-ar^-\-ar^. 

We  next  multiply  each  term  of  the  above  equation  by  r,  and 
we  have 

(2.)  Sr=ar-\-ar'^-{-ar^-Yar'^-{-ar^-{-ar^, 

By  subtracting  the  first  equation  from  the  second,  we  have 
Sr — S=ar^ — a. 

Dividing  by  r — 1,  we  have  the  formula  for  finding  the  sum 

of  the  series 

^     ar^—a        af'—a  (?'"— 1) 

iS= =-,  or  ^r-,  or  a =— . 

r — 1  r — 1  r — 1 

If  the  ratio  is  less  than  a  unit,  we  transpose  the  terms,  thus : 

^     a—ar"^        a—ar""  (1— r") 

6=^j ,  or  — ,  =  a— . 

1 — r  1 — ?•  1 — r 


GEOMETRICAL    PROGRESSION.  237 

270i  The  index  of  the  ratio  is  always  equal  to  the  number  of 
terms. 

By  the  above  formulas,  we  have  a  method  for  finding  the  sum 
of  the  series  without  the  last  term,  which  may  be  expressed  by 
the  following 

Rule.  Raise  the  ratio  to  a  power  wTiose  exponent  is  equal 
to  the  number  of  terms  ;  inultiply  this  power  by  the  first  term, 
find  the  difference  between  this  product  and  the  first  terra,  and 
divide  this  remainder  by  the  difference  between  the  ratio  and 
unity. 

If  we  substitute  the  value  of  L  as  found  in  Art.  274,  we 
shall  have 

^     Lr — a 

o= — . 

r — 1 

A  rule  for  this  formula  would  be  the  same  as  in  Art.  278. 

13.  If  the  first  term  is  7,  the  ratio  3,  and  the  number  of 
terms  5,  what  is  the  sum  of  the  series  ? 

SJ'-^JJ^^^UI.    Ans. 
r — 1  o — 1 

14.  If  the  first  term  is  9,  the  ratio  f ,  and  the  number  of 
terms  4,  what  is  the  sum  of  the  series  ? 

„     a—ar''     9— (9x(^)^      .  , 

15.  If  the  first  term  is  144,  the  ratio  1.06,  and  the  number 
of  terms  4,  what  is  the  sum  of  the  series  ?  Ans.  629.945. 

16.  If  the  first  term  is  9,  the  ratio  |-,  the  number  of  terms  6, 
what  is  the  sum  of  the  series  ?  Am.  llJ-§f^. 

17.  If  the  first  term  is  a,  the  ratio  r,  and  the  number  of 
terms  n,  required  the  sum  of  the  series. 

.         ar^—a     «(?•"— 1) 
r — 1  r — 1 

18.  If  the  first  term  is  1,  the  ratio  2,  and  the  number  of 
terms  7,  what  is  the  sum  of  the  series  ?  Am.  127. 


238  ALGEBKA. 

19.  If  the  first  term  is  5,  the  ratio  10,  and  the  number  of 
terms  7,  what  is  the  sum  of  the  series  ?  Aiis.  5555555. 

20.  If  the  first  term  is  4,  the  ratio  ^,  and  the  number  of 
terms  5,  what  is  the  sum  of  the  series  ?  Arts.  5f^. 

21.  If  the  first  term  is  5,  the  ratio  -i,  and  the  number  of 
terras  5,  what  is  the  sum  of  the  series  ?  Ans.  Gy^^. 

22.  A  gentleman  agreed  with  another  to  board  him  for  9 
days ;  he  was  to  pay  3  cents  for  the  first  day's  board,  9  cents 
for  the  second  day,  27  cents  for  the  third  day,  and  so  on,  in  this 
ratio.  What  was  the  amount  of  the  bill  for  the  gentleman's 
board?  Ans.  $295.23. 

To  find  L,  r,  and  a,  from  the  following  equation. 

r — 1 

MultijDlying  by  r — 1,  Sr — S=L7' — a. 

Resolving  into  factors,  S{r — l)=Lr — a. 

Transposition,  Lr=S{r—l)-\-a. 

Division,  L= . 

?• 

To  find  r  from  the  above  equation. 

r — I 

Multiplying  by  r — 1,  Sj' — S=Lr — a. 

Transposing,  Sr — Lr=S — a. 

Dividing  by  S — L,  ^=  c — f 

To  find  a  from  the  above  equation. 

r — 1  * 

Multiplying  by  r — 1,  Sr — S=Lr — a. 

Transposing,  a=Lr — (r — 1)S. 

23.  If  the  first  term  is  3,  the  ratio  2,  and  the  sum  of  the 
series  93,  what  is  the  last  term  ?  A7is.  48. 


a EOM ETHICAL     PROGRESSION.  239 

24.  Insert  three  geometrical  means  between  l  and  128. 

Am.  2,  8,  32. 

25.  If  tlie  first  term  is  2,  the  last  term  4374,  and  the  number 
of  terms  8,  what  is  the  ratio  ?  Ans.  3. 

26.  If  the  ratio  is  2,  the  number  of  terms  G,  and  the  great- 
est term  128,  what  is  the  least  term  ?  Ans.  4. 

27.  If  the  first  term  is  3^,  the  ratio  f ,  the  number  of  terms 
8,  what  is  the  last  term,  and  what  is  the  sum  of  the  series  ? 

Atis.  Last  term  -fW^^'  ^"^  *^®  ^^^  ^^  series  ^^^ij-^- 

28.  If  the  first  term  is  1,  the  last  term  64,  and  the  number 
of  terms  7,  what  are  the  ratio,  and  the  sum  of  the  series  ? 

A71S.  Ratio,  2  ;  the  sum  of  the  series,  127. 

29.  If  the  last  term  is  64,  the  number  of  terms  7,  and  the 
sum  of  the  series  127,  what  are  the  ratio,  and  the  first  term  ? 

Ans.  Ratio,  2;  the  first  term,  1. 

30.  If  the  first  term  is  2,  the  ratio  4,  and  the  number  of 
terms  12,  what  are  the  last  term,  and  the  sum  of  the  series  ? 

Ans.  Last  term,  8388608;  sum  of  the  series,  11184810. 

31.  The  product  of  three  terms  in  geometrical  progression  i.s 
64,  and  the  sum  of  their  cubes  is  584.  What  are  those  num- 
bers ?  Ans.  2,  4,  8. 

32.  There  are  four  numbers  in  geometrical  progression,  the 
second  of  which  is  less  than  the  fourth  by  24,  and  the  sum  of 
the  extremes  is  to  the  sum  of  the  means  as  7  to  3.  Required 
the  numbers.  A?is.  1,  3,  9,  27. 

33.  It  is  required  to  find  four  numbers  in  geometrical  pro- 
gression, such  that  the  difierence  of  the  two  means  shall  be  14, 
and  the  difference  of  the  extremes  49. 

A?is.  7,  14,  28,  and  56. 
The  following  are  the  two  fundamental  equations  from  wliich 
the  twenty  difi"erent  cases  are  exhibited,  — 

r — 1 
and  which  are  found  in  the  followino; 


240 


No 


13 


14 


ALGEBRA. 

TABLE. 


Giv( 


1 

a,  r,  n 

2 

a,  r,  S 

3 

a^  n^  S 

4 

7',  71,   S 

5 

a,  r,  71 

G 

a,  ?*,  I 

7 

a,  71,  I 

8 

7\  71,  I 

9 

a,  71,  I 

10 

a,  71,  S 

11 

a,  I,  S 

12 

71,  I,  S 

r,  71,  I 


r,  71,  S 


15 

r,  I,  S 

16 

71,  I,   S 

17 

a,  7',  I 

18 

a,  r,  S 

19 

a,  I,  S 

20 

r,  I,  S 

Requir'd. 


Formulae. 


7' 

l{S-l)"-^=a{S—aT-\ 


r"— 1 


S= 


ar — a 


S= 

S= 
S= 


Ir — a 


r-1 


^" ^."— 1 


o;?*" — 7'S=a — S. 
S — a 


a=- 


(r-l)S 

a=- -— . 

r"— 1 

a=lr — (r — 1)S. 

a{S-a)"-'=l{S-l)"-\ 


loo;./ — loo;. a  ,  ^ 

71=^^, \-l. 


log.  r 


71  = 


log.  [a-\-{r—l)S]—\og.  a 


log.  r 
loo;.  ^ — \ocr.a 


71= 


log.{S-a)-\og.{S-l) 


+1- 


71 


log.Z— log.  pr— (r— 1)S] 


log.  ?• 


HARMONICAL     PROGRESSION.  241 

The  last  four  cases  in  the  preceding  table  can  be  performed 
only  by  the  aid  of  logarithms,  as  they  belong  to  exponential  or 
transcendental  equations.  They  will,  therefore,  receive  atten- 
tion in  their  proper  place. 


SECTION   XXII. 

HARMONICAL   PROGRESSION. 

Art.  280i  Three  numbers  are  said  to  be  in  harmonical  pro- 
gression when  the  first  is  to  the  third  as  the  difference  between 
the  first  and  second  is  to  the  difference  between  the  second  and 
third. 

Thus  the  numbers  3,  4,  6,  are  in  harmonical  proportion. 

For  3:6::  4—3  :  6-4. 

Or  fl,  b,  c,  are  in  harmonical  proportion  when 
a  :  c  :  :  h — a  :  c — h. 

Thus,  if  the  length  of  three  strings  of  a  musical  instrument  be 
as  the  numbers  3,  4,  G,  they  will  sound  an  octave  3  to  6,  a  fifth 
2  to  3,  and  a  fourth  3  to  4. 

281.  Four  numbers  arc  in  harmonical  proportion  when  the 
first  is  to  the  fourth  as  the  difference  between  the  first  and 
second  is  to  the  difference  between  the  third  and  fourth.  Thus 
the  numbers  5,  6,  8,  10,  arc  in  harmonic  proportion. 

For  5  :  10  :  :  6-5  :  10-8. 

Strings  of  such  lengths  will  sound  an  octave  5  to  10,  a  sixth 
greater  6  to  10,  a  third  greater  8  to  10,  a  third  less  5  to  8,  and 
a  fourth  6  to  8. 

282.  Any  number  of  quantities,  a,  Z*,  c,  rf,  e,  &c.,  are  in  har- 
monical progression  if  «  :  c  :  :  a — h  :  h — c;  b  :  d  :  :  h — c  : 
c — d  ;  c  :  e  :  :  c — d  :  d — e,  &c. 

21 


242  ALGEBRA. 

283.  The  reciprocal  quantities  in  harmonical  progression  are 
in  arithmetical  progression. 

Thus,  if  a,  b,  c,  d,  e,  &c.,   are  in  harmonical   progression, 

1    1    1    1   1    p  .„  ,     . 

-,  -,  -,  -,  -,  &c.,  will  be  m  arithmetical  progression. 

a  b    c    a  e  ^     ° 


SECTION  XXIII. 

INFINITE   SERIES. 

Art.  284.  An  infinite  decreasing  geometrical  series  is  one 
whose  ratio  is  less  than  unity,  and  the  number  of  whose  terms  is 
infinite. 

To  find  the  sum  of  an  infinite  series  decreasing  in  geometrical 
progression. 

We  have  already  found.  Art.  277,  that  the  sum  of  a  descend- 
ing series  in  geometrical  progression  may  be  ascertained  by  the 
following  formula. 

_,     a — «?•"         ^        a         ar"" 
o=— ,  or  o=: 


1 — r  1 — r     1 — r 

285.  Now,  if  ?'"  be  a  fraction  less  than  a  unit,  it  is  evident 
that  the  greater  the  number  ?z,  the  smaller  will  be  the  quantity 
r".  If,  therefore,  a  great  number  of  terms  of  a  descending 
series  be  taken,  the  quantity  r"  will  be  very  small ;  and,  if  we 
suppose  n  greater  than  any  assignable  number,  then  the  quan- 
tity, or  its  value,  may  be  considered  as  nothing  =  0. 

Hence    the    latter   part   of  the    formula,    — ,  should  be 

1 — r 

omitted,  and  it  will  stand 

a 


Thus,  S=-. 


\-r 

The  rule,  therefore,  for  finding  the  sum  of  the  series,  is  as 
follows  : 

Rule.  Divide  the  first  term  by  the  dijfcrence  between  unity 
arid  tJie  ratio. 


INFINITE     SERIES. 


EXA3IPLES. 


243 


1.  AYhat  is  the  sum  of  the  infinite  series,  1,  ^,  ^,  ^V'  ¥i' 

&c.? 

1        1 


=14.     Atis. 


1 1         2  ^ 

2.  What  is  the  sum  of  1,  ^,  ^i,  |,  &c.,  to  infinity  ?   J.715.  2. 

3.  What  is  the  sum  of  the  series,  8,  |,  ^\,  yf  y,  &e.,  carried 
to  infinity  ?  Aiis.  10. 

4.  Find  the  value  off,  4,  ^,  yV'  <^^'»  *^  infinity.    J.7Z5.  1*. 

5.  Find  the  value  of  4,  1,  ^,  y^g,  &c.,  to  infinity.    A?is.  5i. 

6.  AYhat  is  the  exact  sum  of  1,  yV,  i^^'  <^c-»  ^^  infinity  ? 

^?w.  1^. 

7.  Find  the  exact  value  of  the  circulating  decimal  .444,  &c., 
to  infinity. 

.444,  &c.=^+y^^+y^VTT.  tlie  ratio  being  J^- 

jt         _* 

ICF      TO" _^    VIO :tO *  Atj^ 


1 1  9 

[See  National  Akithmetic,  page  128.] 
8.  What  common  fraction  will  exactly  express  the  value  of 
the  repeating  decimal  .454545,  &c.  ? 

.454545=^^^+y^VW+inJxr*^TjTr>  t^e  ratio  being  y^. 

4  5                 4  5 
Ta^     l^*^ 45    VlOO 4500 5  /I77C 


•^        10  0         loTJ 

9.  What  common  fraction  is  the  exact  value  of  the  decimal 
.571428?  A?2s.  f 

10.  What  common  fraction  is  the  exact  value  of  .857142  ? 

Atis.  f. 

11.  AVhat  is  the  exact  value  of  .53  ? 

•5o=y^  and  TTTTr~riooo  I  iooott?  &c. 


3 

1- 


TD^y     TO'TJ ;?      v/  1  0 30   1     .       5      I       l    8  A-ntf 

TU        TTT 


12.  What  is  the  value  of  .138  ?  A?is.  -^^. 

13.  Find  the  ratio  of  an  infinite  series  whose  first  term  is  8, 
and  the  sum  of  the  series  10.  Atis.  4-. 


244  ALGEBRA. 

14.  Find  the  ratio  of  an  infinite  series  whose  first  term  is  |, 
and  whose  sum  is  IJ-.  Ans.  J-. 

15.  Find  the  first  term  of  an  infinite  progression  of  which 
the  ratio  is  \,  and  the  sum  10.  Ans.  8. 


SECTION   XXIV. 

SIMPLE   INTEREST. 

Art.  286.  Interest  is  the  compensation  which  the  borrower 
makes  to  the  lender  for  the  use  of  a  certain  sum  of  money  for  a 
given  time. 

Pi'incipal  is  the  sum  lent. 

Rate  per  cent,  is  the  sum  agreed  on  for  the  loan  of  $1,  or 
$100,  for  one  year. 

Amownt  is  the  sum  of  the  interest  and  principal. 
Legal  interest  is  the  rate  per  cent,  established  by  law. 
Let  p  =  principal. 

r  =  rate  per  cent.,  written  in  hundredths. 
t  =  time  in  years. 
a  =  amount. 
i  or  a — p  =  interest  for  the  given  time. 
Hence,  if  r  be  the  interest  of  one  dollar  for  one  year,  it  is 
evident  that  the  interest  of  ^  dollars  will  be  p  times  r=pr. 

And  ifpr  be  the  interest  of  p  dollars  for  one  year,  it  is  cer- 
tain that  for  t  years  it  will  be  t  times  as  much,  =  ^^ifr,  and  that 
p~\-ptr  will  be  the  amount,  and  i  or  a — p  will  be  the  interest. 

287.  Hence,  having  the  principal,  rate  per  cent.,  and  time 
given,  to  find  the  interest  and  amount,  we  have  the  following 
formulae  : 

Formula  for  the  interest, 

i=ptr. 
Formula  for  the  amount, 

a=p-\-ptr. 


SIMPLE    INTEREST.  245 

From  the  preceding  formulae  we  have,  for  finding  the  interest 
and  amount,  the  following 

KuLE.  Multiply  the  'princiiml  by  the  rate  per  cent.,  considered 
as  a  decimal,  and  this  product  by  the  time  in  years,  and  the  result 
is  the  interest. 

If  there  are  months  and  days,  let  the  months  be  considered  as 
fractions  of  a  year,  and  the  days  as  fractions  of  a  month. 

By  adding  the  interest  to  the  principal,  zve  have  the  amount. 

[See  National  Akithjietic,  page  164.] 

EXAMPLES. 

1.  What  is  the  interest  of  $740  for  4  years,  at  6  per  cent.  ? 

i=;??r=740x.06x4==$177.60.     Ans. 

2.  What  is  the  interest  of  $380  for  10  years,  at  5  per  cent.  ? 

Alls.  $190. 

8.  What  is  the  interest  of  $890.75  for  3  years,  6  months,  at  8 
percent.?  ^725.  $249.41. 

4.  What  is  the  interest  of  $17.18  for  5  years,  2  months,  10 
days,  at  4i  per  cent.  ?  Ans.  $4.02. 

5.  What  is  the  amount  of  $144  for  3  years,  at  8  per  cent.  ? 

a=;7+;?r^=144+(144x.08x3)=$178.56.     Ans. 

6.  What  is  the  amount  of  $800  for  6  years,  1  month,  12  days, 
at  6  per  cent.  ?  Ans.  $1093.60. 

7.  What  is  the  amount  of  $670.18  for  3  years,  7  months,  20 
days,  at  9  per  cent.  ?  Ans.  $889.66. 

288.  Having  the  amount,  time,  and  rate  per  cent,  given,  to 
find  the  principal. 

By  transposing,  &c.,  the  last  equation,  we  have 

a 

From  which  we  have  the  following 

KuLE.  Mzdtiply  the  time  by  the  rate  per  cent.,  and  add  1  to 
the  product ;  ivith  this  sum  divide  the  amount,  and  the  quotient 
is  the  principal. 

21^' 


246  ALGEBRA. 

8.  Keceived  $472  for  a  certain  sum  that  had  been  on  interest, 
at  6  per  cent.,  for  3  years.     What  was  the  sum  lent  ? 

a  472  ^Ar^r^  A 

^     l+tr     1+(3X.06) 

9.  What  principal  will  amount  to  $570  in  10  years,  at  5  per 
cent.?  Am.  $380. 

10.  What  principal  will  amount  to  $1140.16  in  3  years,  6 
months,  at  8  per  cent.  ?  Ans.  $890.75. 

11.  Lent  a  certain  sum  for  5  years,  2  months,  10  days,  at  4^ 
per  cent.,  and  received  interest  and  principal  $21.20  ;  what  was 
the  sum  lent?  tI?!^.  $17.18. 

12.  My  friend  borrowed  of  me  a  certain  sum,  which  he  kept 
3  years,  and  for  which  I  charged  him  8  per  cent.,  and  received 
interest  and  principal  $178.56.     What  was  the  sum  I  lent  him  ? 

Am.  $144. 

13.  Keceived  as  interest  and  principal  $889.66  from  a  friend 
to  whom  I  had  loaned  a  certain  sum  for  3  years,  7  months,  and 
20  days,  at  9  per  cent.  What  was  the  consideration  of  his 
note?  Am.  $670.18. 

289.  Having  the  amount,  principal,  and  rate  per  cent,  given, 
to  find  the  time. 

By  transposing  and  reducing  the  last  equation,  we  have  the 
following  formula  for  finding  the  time,  t. 

rp       rp 
From  the  above  formula  we  have  the  following 

Rule.  Divide  the  interest  hy  the  jrroduct  of  the  principal 
multiplied  by  the  rate  per  ce7it.,  and  the  quotient  is  the  time. 

[See  National  Arithmetic,  page  181.] 

14.  How  long  will  it  require  $300  to  amount  to  $372,  at  6 
per  cent.  ? 

T   .  .     «-^     372-300     ^ 

Let  t= =-7T7: — Frr77:=4  years.     Ans. 

rp        .06X300        -^ 

15.  In  what  time  will  $380  amount  to  $570,  at  5  per  cent.  ? 

Ans.  10  vears. 


SIMPLE     INTEREST.  247 

16.  Lent,  at  8  per  cent.,  $890.75,  for  which  I  received 
$1140.16  ;  for  how  long  time  was  the  money  lent  ? 

A?is.  3  years,  6  months. 

17.  For  $17.18,  which  was  loaned  at  4J-  per  cent.,  there 
was  received  $21.20.     For  how  long  time  had  it  been  lent  ? 

Atis.  5  years,  2  months,  10  days. 

18.  The  interest  and  principal,  on  a  certain  sum,  at  9  per  cent., 
are  $889.66;  and  the  interest  is  $670.18  less  than  the  amount. 
How  long  was  the  money  at  interest  ? 

A71S.  3  years,  7  months,  20  days. 

19.  A  has  B's  note,  dated  January  1,  1851,  for  $320,  at  9 
per  cent.     When  will  the  note  amount  to  $353.60  ? 

A?is.  March  1,  1852. 

290«  Having  the  principal,  interest  and  time  given,  to  find 
the  rate  per  cent. 

By  transposing  the  last  formula,  we  obtain  the  following  for 

finding  r,  the  rate  per  cent.     Thus, 

a — p         i 
"  or 


pt  jjt 

The  pupil  will  perceive  that  the  amount  is  known  when  the 
interest  and  principal  are  given. 

What  is  the  rate  per  cent,  for  $300,  that  it  shall  amount  to 

$372  in  4  years? 

a-p     372-300       „.        ^ 

r= =-— — — --=  .06,  or  o  per  cent. 

pt         300x4  ^ 

Hence  we  deduce  the  following 

Rule.     Divide   the  interest  by  the  product  of  the  principal 
inulliplied  hy  the  time,  and  the  quotient  is  the  rate  per  cent. 

20.  If  $380  amount  to   $570  in  ten  years,  what  is  the  rate 
per  cent.  ?  Aiis,  5  per  cent. 

21.  Lent  $890.75,  for  3  years,  6  months,  and  received  for  the 
amount  $1140.16.     'WTiat  was  the  rate  per  cent.  ? 

Ans.  8  per  cent. 

22.  If  $17.18  amount  to  $21.20  in  5  years,  2  months,  and  10 
days,  what  is  the  rate  per  cent.  ?  Ans.  4:^  per  cent 


248  ALGEBRA. 

23.  If  the  interest  of  $670.18  for  3  years,  7  montlis,  and  20 
days,  be  $219.48,  what  is  the  rate  per  cent.  ? 

Ans.  9  per  cent. 

24.  John  Smith,  Jr.,  gave  me  his  note,  dated  January  1, 
1848,  for  $144  :  but  he  having  been  unfortunate  in  business,  I 
agreed,  May  7,  1851,  to  give  him  up  his  note  for  S153.64.8. 
What  per  cent,  did  I  receive  ?  Ans.  2  per  cent. 

25.  My  tailor  informs  me  that  my  "  freedom  suit "  will  re- 
quire 7^  square  yards  of  cloth ;  but  the  cloth  I  am  about  to 
purchase  will  shrink  5  per  cent,  in  width,  and  4  per  cent,  in 
length,  and  the  cloth  is  60  inches  wide.  How  many  yards  must 
I  purchase  ?  Ans.  4  yards,  38{|  inches. 


SECTION  XXV.  . 

DISCOUNT   AT  SIMPLE   INTEREST. 

Art.  291 .  Discount  is  an  allowance  for  the  payment  of  any 
sum  of  money  before  it  becomes  due,  and  is  the  difference 
between  that  sum  and  its  present  worth. 

The  present  worth  of  any  sum  due  some  time  hence  is  such  a 
sum  as,  if  put  at  interest,  would,  in  the  time  for  which  the  dis- 
count is  to  be  made,  amount  to  the  sum  then  due. 

To  find  the  worth  of  any  sum  due  at  any  time  hence : 
Let  S  =  the  sum  due. 

p  =  the  present  worth. 
t  =  the  time  in  years. 

r  =  the  rate  per  cent,  considered  as  so  many  hun- 
dredths. 

We  have  before  shown,  in  Art.  287,  that  a=p-^ptr. 


DISCOUNT     AT     SIMPLE     INTEREST.  249 

We  now  substitute  S  for  a,  and  consider  p  to  represent  the 
present  worth  ;  and,  by  transposing  the  equation,  find 

S 

from  which  we  deduce  the  following 

Rule.  Multiply  the  time  ly  the  rate  per  cent.,  add  1  to  the 
product,  and  divide  the  sum  on  vjhich  the  discount  is  to  he  taken 
by  this  su.m,  and  the  quotient  is  the  present  vjorth. 

If  the  present  worth  is  taken  from  the  sum  due,  the  re- 
mainder is  the  discount. 

[See  National  Arithmetic,  page  187.] 

1.  What  is  the  present  value  of  $500,  due  4  years  hence,  at 
6  per  cent.  ? 

;?=,-^-=--4^^-=$403.22+.     Ans. 
^     l-\-tr      1  +  (4X.0G)  ^ 

By  transposing  the  quantities  in  the  above  formula,  we  may 
obtain  the  values  of  5,  t,  and  r. 

2.  What  is  the  present  worth  of  $372,  due  4  years  hence,  at 
6  per  cent.  ?  Ans.  $300. 

3.  What  is  the  present  worth  of  8133.20,  due  20  months 
hence,  at  8|-  per  cent.  ?  An^.  $117.09, 

4.  What  is  the  discount  on  $21.20,  due  5  years,  2  months, 
10  days  hence,  at  4i  per  cent.  ?  Ans.  $4.02. 

5.  A  has  B's  note,  dated  January  1,  1851,  for  $353.60,  to  be 
paid  March  1,  1852,  without  interest.  What  was  the  value  of 
this  note  at  the  time  it  was  given,  if  9  per  cent,  discount  is 
allowed?  Ans.  $320. 

6.  Which  is  worth  the  most,  A's  note  for  $144,  due  10  years 
hence,  at  6  per  cent.,  or  B's  note  for  $176.40,  due  8  years 
hence,  at  12  per  cent.  ?  Ans. 

7.  A  legacy  of  $1725  is  due  one  year  hence.  What  is  its 
present  value,  at  15  per  cent.  ?  Am.  $1500. 

8.  James  Brown  has  S.  Smith's  note  for   S162,  payable    6 


250  ALGEBRA. 

months  hence ;  but  Brown,  being  obliged  to  raise  money,  sold 
the  note  for  $150.     What  per  cent,  did  he  allow  ? 

Ans.  16  per  cent. 

9.  Bought  a  farm  for  $590,  for  which  I  was  to  pay  in  a  cer- 
tain time,  without  interest ;  but,  by  making  prompt  payment,  I 
was  allowed  a  discount  of  6  per  cent,  for  the  whole  time,  and 
paid  only  $500.     How  long  was  the  time  allowed  for  payment  ? 

A71S.  3  years. 

10.  Bought  a  horse  for  $200,  and  gave  my  note,  payable  in 
60  days.  What  ready  money,  at  15  per  cent.,  will  discharge 
the  debt?  Ans.  $195.12+. 

11.  What  is  the  present  worth  of  $1827,  due  100  years 
hence,  at  6  per  cent.  ?  Ans.  $261. 


SECTION  XXVI. 

PARTNERSHIP,    OR   COMPANY  BUSINESS. 

Art.  292i  Partnership  is  the  association  of  two  or  more 
persons  in  business,  with  an  agreement  to  share  the  profits  and 
losses  in  proportion  to  the  amount  of  the  capital  stock  con- 
tributed by  each. 

EXAMPLES. 

1.  Three  men.  A,  B  and  C,  enter  into  partnership  for  two 
years,  with  a  capital  of  $1600.  A  puts  into  the  firm  $300,  B 
$500,  and  C  $800.  They  gain  $320.  What  is  each  man's 
share  of  the  gain  ? 

Let  a:  =  A's  gain. 

Then,  as  each  man's  share  of  the  gain  will  be  in  proportion  to 
his  stock. 

And  -— -  =  B's  gain. 

8a; 

—  =  C's  gain. 


PARTNERSHIP,    OR    COMPANY   BUSINESS.  251 

And  x+^-i-^  =  S320. 

o        o 

Sx+bx-\-Sx  =  960. 

16a;  =  960. 

X  =    60  =  A's  gain. 

::^  =  100  =  B's  gain. 

o 

^  =  160  =r  C's  gain. 
o 

VERIFICATION. 

60-f-100-|-160=$320. 

Or,  let  m,  n,  and  'p  represent  A,  B,  and  C's  stock,  and  a  the 
sum  gained. 

Also,  let  :6-  =  A's  gain. 

Then,  it  is  evident  that  each  man  must  receive  according  to 
his  capital. 

That  is,  as  A's  stock  is  to  his  gain,  so  will  B's  stock  be  to  his 

gain,  &c. 

nx 
Therefore,  m  :  x  :  '.  n  :  —  =  B  s. 

m 

px        ^ 

And  m  :  X  :  :  p  :    —  =  C  s. 

m 

Then,  x-] \-^  =  a. 

m      m 

And  7nx-\-nx-\-px^=am. 

^,       ,  am  320x300  „,.„    ,,       . 

Therefore,  a:=— —=—-—4^--— -=$00,  A's  gam. 

m-\-7i-{-p      oUO-f-oOO-j-oOO 

Then,  by  the  principle  above  stated, 

a7?i  an  320x500  ^, 

«i  • •  •  77  • = =8100,  B  s  gam. 

'''•ra-\-n-\-p'        m+n-\-p     300+500+800     ^        '         ^ 

And, 
am  ap  320x800  „^  ^, 

m-\-7i-{-p  m-\-n-\-p     oUU+oUU+oUU 


252  ALGEBRA. 

•VERIFiaA.TION. 

am  an  ap      "     [m~{-n-\--p)a 


=a=$320. 


m-\-n-{-p     7n-\-n-\-p     m-\-7i-\-p       m-\-n-\-p 

Therefore,  to  find  the  gain  or  loss  on  any  man's  stock,  we 
deduce  from  the  above  formulae  the  following 

Rule.     Multiply  the  whole   gain  by  each  man^s  stock,  and 
divide  the  product  by  the  whole  stock. 

293.  Having  each  man's  gain,  and  the  amount  of  stock  given, 
to  find  each  man's  share  in  the  stock. 

.  2.  A,  B,  and  C,  while  in  trade,  gained  as  follows.  A  gained 
$50,  B  $70,  and  0  $90.  The  amount  of  their  stock  in  trade 
was  $4200.     What  was  the  amount  of  each  man's  stock  ? 

It  is  evident  that  each  man's  stock  was  in  proportion  to  his 
sain. 


Let         a;=A's  stock. 

Ix 
Then    ~=B's  stock. 
5 

9a; 
And     --=C's  stock. 
5 

Therefore,  :.+-^4-:^=4200. 

0         0 

5:^_|_7x+92-=21000. 

21:r=21000. 

a:=1000.    A's 

stock. 

5=1400.    B's  stock. 
5 

^=1800.    C's  stock. 
5 


4200.    Proof. 

If  we  change  the  symbols  of  the  first  question,  putting  m,  7^, 
and  p,  for  the  gain  of  each  man  respectively,  and  a  for  the  stock, 
wc  obtain  the  followinor  formuhx)  for  tindinji;  •  the  amount  of 
each  man's  stock : 


PARTNERSHIP,     OR     COMPANY    BUSINESS.  253 

ma  50x4200 


m-\-n-{-p     50+70-1-90 
na  70x4200 


m-\-n-]-p     50-j-70-f90 
pa  90x4200 


=  $1000.     A's  stock. 
$1400.     B's  stock. 


=  $1800.     C's  stock. 


m-^n+p     504-70-1-90 

Hence,  for  finding  each  man's  stock,  we  have  the  following 

Rule.  Midtiply  the  whole  stock  hy  each  man^s  gain,  and 
divide  the  product  hy  the  lahole  gain. 

3.  Two  men,  M  and  N,  engaged  in  trade.  M  put  in  $500, 
and  N  $750.     They  gained  $120.     What  is  each  man's  gain  ? 

Ans.  M  gained  $48,  N  gained  $72. 

4.  Q  and  X  hired  a  field  for  $120,  which  they  used  for  a 
pasture.  Q  put  in  11  cows,  and  X  15  cows.  What  sura  should 
each  man  pay  ?        Ans.  Q  pays  $50.76if ,  X  pays  $69.23^5. 

5.  A  and  B  purchased  a  factory  for  $17,000.  A  paid  $10,000, 
and  B  the  remainder.  They  gained  $1500.  What  sum  should 
each  receive  ?  Ans.  A  $882-iV,  B  $0171^. 

6.  A,  B,  and  C  engaged  in  trade,  with  a  capital  of  $6000. 
They  gained  $240.  A's  share  of  the  gain  was  $100,  B's  $80, 
and  C's  $60.     What  part  of  the  stock  did  each  own  ? 

Ans.  A  $2500,  B  $2000,  and  C  $1500. 

7.  A,  B,  and  C  hire  a  pasture  for  the  season  for  $100.  A 
put  in  5  horses,  B  7  oxen,  and  C  9  cows.  Two  horses  eat  as 
much  as  3  oxen,  and  4  oxen  eat  as  much  as  5  cows.  What  part 
of  the  expense  must  each  pay  ?  Ans.  A  pays  $34.56^Y_^  B 
pays  $32.25i^^,  and  C  pays  $33.17f  if . 

8.  Three  men.  A,  B,  and  C,  agreed  to  reap  a  field  that  was 
40  rods  square  for  $32.  A  reaped  a  part  that  was  25  rods 
square,  B  reaped  400  square  rods,  and  C  the  remainder.  What 
sum  did  each  receive?  Ans.  A  $12.50,  B  $8,  C  $11.50. 

PARTNERSHIP  ON  TIME,  OR  DOUBLE  FELLOWSHIP. 

9.  A,  B,  and  C  engaged  in  trade.  A  put  in  $2000  for  4 
months,  B  put  in  $3000  for  8  months,  and  C  put  in  $4000  for 

22 


254 


ALGEBRA. 


12  months.     They  gained  $780.     What  is  each  man's  share  of 
the  gain  ? 

Let  w,  n,  p,  represent  each  man's  stock,  a  the  whole  gain, 
and  t,  f,  t'\  the  time  each  man's  stock  was  in  trade.  It  is 
evident  that  each  man's  stock  gains  not  only  in  proportion  to  its 
sum,  but  also  in  proportion  to  the  time  it  is  in  trade.  For 
$2000  will  gain  four  times  as  much  in  four  months  as  it  would 
in  one  month,  and  $2000  for  four  months  is  the  same  as  $8000 
for  one  month.  We  must,  therefore,  multiply  each  man's  stock 
by  the  time  it  was  in  trade.  It  is  therefore  evident,  that  as  A's 
gain  is  to  B's  gain,  as  A's  stock  multiplied  by  his  time  is  to  B's 
stock  multiplied  by  his  time,  &c. 

Let  a:,  ?/,  z  =  A,  B,  C's  gain  respectively. 

X  :  y  :  :  mt  :  nt', 

7it'x 
y= — —  =  B  s  gam. 


Then 

Multiplying  extremes,  &c., 

And 

Multiplying  extremes,  &c., 


2r= 


mt 

z  :  ; 
pt"x 


And 

Multiplying  by  mt 

Therefore, 
mta 


ntx 


X- 


lilt 
pf'x 


mt  :  pt". 
=  C's  gain. 


•a. 


mt       mt 
mtx-\-nfx  -\-pf'xz=mta. 


mt-{-7it'+pt' 
But 


2000X4X780 ^ 

2000x4+3000x8+4000x12      *  A'^gaia 


y- 


nt'x 

mt ' 


.     _  ,         ...  nt'  mta  nt'a 

And  by  substitution,  2/=-,X,;;^^q:^,q:^=  ^^^_^^^,,_^^,= 

ggggX8x780 

2000X4+3000X8+4000X12  ^ 

plf'x 


And 

And  by  substitution, 


mt 


PAllTNERSHIP     ON     TIME.  255 


pf  mta  pf'a 


mt      mt-\-nt' -\-pt"     mt+nt'-{-pt" 

4000X12X780  __  . 

2000X4>3000X«+4000X"12"  ^  O  s  gam. 

The  above  equations,  by  dividing  the  numerators  each  into 
two  factors,  may  be  expressed  by  the  following  proportions : 
7nt-\-nf -\-pf'  :  mt  :  :  a  :  x. 
mt-{-nt'-{-pt"  :  nf  :  :  a  :  y. 
mt-\-nt'-\-pt"  :  jyt"  :  '.  a  :  z. 
Hence  the  following  arithmetical 

E-ULE.  Multiply  each  marCs  stock  by  the  time  it  was  continuea 
in  trade,  aiid  then  say,  As  the  sum  of  all  the  products  is  to  each 
man's  product,  so  is  the  whole  gain  or  loss  to  each  man's  gain 
or  loss.  [See  National  Arithmetic,  Sec.  LVI.] 

10.  A  commenced  business  January  1,  1850,  with  a  capital 
of  83000.  May  1,  1850,  he  took  B  into  partnership,  with  a 
capital  of  $4000.  January  1,  1851,  they  had  gained  $340. 
What  was  each  man's  share  of  the  gain  ? 

Ans.  A's  gain  $180,  B's  gain  $160. 

11.  A,  B,  and  C  traded  in  company.  A  put  in  $300  for  10 
months,  B  put  in  $400  for  8  months,  and  C  put  in  $600  for  2 
months.     They  gained  $120.     What  is  the  gain  of  each  ? 

Ans.  A's  gain  $48.64^1,  B's  $51.89/^,  C's  $19.45f  f. 

12.  Three  men,  A,  B,  and  C,  hire  a  pasture  in  common,  for 
which  they  are  to  pay  $76.80.  A  put  in  24  oxen  for  12  weeks. 
B  put  in  25  oxen  for  12  weeks,  and  C  put  in  30  oxen  for  6 
weeks.     What  sum  ought  each  to  pay  ? 

Am.  A  $28.80,  B  $30,  C  $18. 

13.  John  Jones  hired  a  house  for  one  year  for  $500,  with  the 
privilege  of  admitting  two  more  families  if  he  pleased,  with  the 
understanding  that  all  the  occupants  should  have  equal  priv- 
ileges in  the  house.  At  the  end  of  three  months  he  took  in 
John  Smith,  and  at  the  end  of  9  months  Richard  Roe.  What 
share  of  the  rent  should  each  pay? 

^7?.^.  Jones  $291f ,  Smith  $166|,  Roe  $41f . 


256  ALGEBRA. 

14.  Two  men,  A  and  B,  hired  a  coach  in  Boston  to  go  to 
Worcester,  the  distance  being  42  miles,  for  $20,  with  the  privi 
lege  of  taking  in  two  persons  more.  Having  rode  80  miles, 
they  take  in  C  ;  and  on  their  return  from  Worcester,  when 
within  20  miles  of  Boston,  they  take  in  D.  What  ought  each 
man  to  pay  for  his  accommodation  in  the  coach  ? 

Am.  A  $7.46^1^,  B  $7.46^f  j,  C  $3.88f||,  D  $1.19^l2_ 

15.  A  and  B  engage  in  trade.  A  puts  in  a  dollars  for  b 
months,  B  puts  in  c  dollars  for  d  months,  and  they  gain  e  dol- 
lars.    What  share  of  the  gain  shall  each  receive  ? 

A71S.     —r-, — 7  As  gain.     ——■ — -  B  s  gain. 
ab-\-cd         ^  ab^cd         ^ 

16.  A,  B,  and  C  engage  in  trade,  with  a  capital  of  $1911. 
A's  money  was  in  the  firm  3  months,  B's  5  months,  and  C's  7 
months.  They  gained  $117,  which  was  so  divided  as  that  the 
^  of  A's  gain  was  equal  to  -^  of  B's  and  ^  of  C's  gain.  What 
was  each  man's  stock  and  gain  ? 

(  A's  stock  $693^2_VV'  ^'s  S623|ffi|,  and  C's  $594/^3^^. 
\  A's  gain  $26,  B's  gain  $39,  and  C's  gain  $52. 

17.  If  12  oxen  eat  3^  acres  of  grass  in  4  weeks,  and  21  oxen 
eat  10  acres  in  9  weeks,  how  many  acres  would  36  oxen  eat 
in  18  weeks,  the  grass  to  be  growing  uniformly  ? 

Atis.  24  acres. 

18.  Three  men  engage  in  partnership,  for  20  months;  A,  at 
first,  put  into  the  firm  $4000,  and  at  the  end  of  4  months  he  put 
in  $500  more ;  but,  at  the  end  of  16  months,  he  took  out  $1000. 
B,  at  first,  put  in  $3000,  but  at  the  end  of  10  months  he  took 
out  $1500,  and  at  the  end  of  14  months  he  put  in  $3000.  C,  at 
first,  put  in  $2000,  and  at  the  end  of  6  months  he  put  in  $2000 
more,  and  at  the  end  of  14  months  he  put  in  $2000  more ;  but, 
at  the  end  of  16,  he  took  out  $1500.  They  had  gained,  by 
trade,  $4420.     What  is  each  man's  share  of  the  gain  ? 

A71S.  A's  gain,  $1680  ;  B's  gain,  $1260  ;  C's  gain,  $1480.  • 


INDETERMINATE     ANALYSIS.  257 

SECTION   XXVII. 

INDETERMINATE   ANALYSIS. 

Art.  294,  In  the  common  rules  of  Algebra,  such  questions 
are  usually  proposed  as  require  some  certain  or  definite  answer  ; 
in  which  case,  it  is  necessary  that  there  should  be  as  many  inde- 
pendent equations,  expressing  their  conditions,  as  there  are 
unknown  quantities  to  be  determined ;  otherwise  the  problem 
would  not  be  limited. 

But,  in  other  branches  of  the  science,  questions  frequently 
arise  that  involve  a  greater  number  of  unknown  quantities  than 
there  are  equations  to  express  them ;  in  which  instance,  they  are 
called  indetermiTiate^  or  unlimited  problems,  being  such  as 
commonly  admit  of  an  indefinite  number  of  solutions  ;  although, 
when  the  question  is  proposed  in  integers,  and  the  answers  are 
required  only  in  whole  positive  numbers,  they  are  in  some 
cases  confined  within  certain  limits,  and  in  others  the  problem 
may  become  impossible. 

Note. —  The  rule  of  Alligation  belongs  to  Indeterminate  Analysis.  See 
the  Author's  National  Akitumetic,  page  275. 

EXAMPLES. 

1.  Let  5a;+3?/=49. 

It  is  required  to  solve  the  equation,  and  find  all  the  integral 
and  positive  values  of  x  and  y  which  are  possible. 
(1.)  By  transposition,  3?/=49 — bx. 
(2.)  Dividing  as  far  as  possible, 
2/=lo — X- 


a 

By  changing  the  fraction,  for  the  sake  of  convenience,  to  a 
positive  quantity, 

(3)  2^=16-20;+^. 

Since  we  consider  only  the  integral  values  of  y,  the  fraction 
must  be  a  whole  number. 
22=^ 


258  ALGEBRA. 

Let  ?z=tliat  number. 

Then        7i=^±l. 
o 

x=Sn — 1. 
Substituting  this  value  ofx  in  (3), 

We  have,  y=lQ—2{Sn—l)+n. 

Or,  2^=18 — K>n. 

We  have  now  the  values  of  x  and  y  in  the  terms  of  n,  which 
must  be  whole  numbers. 

By  trying  various  values  for  n,  we  shall  find  all  the  possible 
values  of  x  and  y. 

Let  72=1,  and  x=  2,  and  y=13. 

71=2,    "   x=  5,    "   y=S. 

w=3,    "   x=  8,    "    7j=d. 

n—4:,    "    a:=ll,    "    y=—2. 

This  last  value  of  y,  being  negative,  is  not  allowed  by  the 
conditions  of  the  question. 

The  equation,  therefore,  admits  of  only  three  sets  of  answers. 

2.  How  can  $100  be  paid  with  100  pieces,  using  eagles, 
dollars,  and  "  nine-pences,"  each  of  the  latter  equalling  one- 
eighth  of  a  dollar  ? 

Let  a:=eagles,  2/= dollars,  z=nine-pences. 

(1)  Then,  2+?/+z=100. 

(2)  And  10.i'+7/+|=100. 

(3)  Multiplying  (2)  by  8,        SQx+Sy-{-z^SQO. 

(4)  Subtracting  (1)  from  (3),    '     19x-\-7y=7Q0. 

(5)  Transposing,  77/=700— 7U2\ 

(6)  Dividing,  2/=100— 12:c-f-^. 

(7)  Let  7i=~. 


79n 


INDETERMINATE     ANALYSIS.  259 

(8)  7n=bx. 

In 

(9)  ^=T- 

(10)  By  substitution,        7/=100- 

Let  n  =  b,  it  being  the  smallest  number  that  will  give  an 
integral  value  to  x  ;  and  we  find  x=7,  and  y=21,  and  z=^72. 

Again,  let  7i  =  10,  the  next  smallest  number  that  will  make 
X  a  positive  whole  number,  and  we  find  a:=14,  and  y  a  negative 
quantity ;  and  so  with  every  value  of  7i  that  can  be  assumed, 
except  5.  The  question,  then,  admits  of  but  one  answer ;  that 
is,  7  eagles,  21  dollars,  and  72  nine-pences. 

The  answer  might  have  been  obtained  by  eliminating  y,  instead 
of  z. 

(1)  Thus,  x+y-\-z=100. 

(2)  And  102--|-^+|=100. 


(3) 

Subtracting 

(1) 

from 

(2), 

9x- 

-T=«- 

(4) 

Multiplying, 

7z=72x. 

(5) 

Dividing, 

z=10x-^^. 

(6) 

Let 

• 

2x 

G) 

Multiplying, 

77i=2x. 

(8) 

Dividing, 

7n 
x=—. 

Let  ?z=2,  it  being  the  least  number  that  will  make  x  a  wholo 
number,  and  Xz=.7,  and  z=72,  and  7/=21. 

If  we  suppose  ?z=4,  it  being  the  next  larger  number  that  will 
make  x  an  entire  number,  then  a:=14,  and  £:=144,  which  is 
impossible,  by  the  conditions  of  the  question.  It  is,  therefore, 
certain  that  no  numbers  but  7,  21  and  72,  are  correct. 


260  ALGEBRA. 


3.  Let       x-{-     y-\-     z=  41  )  to  find  all  the  integral  and  pos- 
And24:r-}-197/-j-102=741  )      itive  values  of  a:,  y,  and  z. 

(1)  Conditions,  x-\-y-\-z=4:l. 

(2)  And  24^-\-197j-\-10z=74:l.       (^ 

(3)  Transposing  (1),  z=41—x—y.       y 

(4)  Transposing  &c.  (2),  z= -. 

(5)  Values  of  (3)  and  (4),  4:l—x—y=^-^^—^:^^. 

(6)  Multiplying,         410— 10:c— 10?/=741— 24a:— 1%. 

(7)  Reducing,  %+14a;=331. 

(8)  Transposing  and  dividing,  y= =SQ—x-{-— — —- 

Changing  the  signs  in  the  last  term,  so  as  to  make 

(9)  X  positive,  y=^Q—x — . 

(10)  Let  St-^=^- 

(11)  Multiplying,  bx—7=9n. 

(12)  Dividing,  a:=^^i^. 

o 

(13)  Substituting  this  for  the  value  of  x  in  the  equation  (9), 

we  have  9^  17 

y=oQ^ n. 

(14)  Multiplying,         5y=1^0—9n—7—^n. 

(15)  Reducing,  by=17^—14:n. 

Let  n=  2,  then  2/=29,  and  x=:  5,  and  z=  7. 
n=  7,    "     ?/=15,    "   xz=14,    "    z=12. 
n=12,    "     7/=  1,    "   x=2S,    "    2:=17. 
Another  solution  of  the  above  question  : 

(1)  Let  x-]-y-{-z=  41. 

(2)  And  242;+19?/-|-10^=741. 


INDETERMINATE     ANALYSIS.  261 


(3) 

Eliminating  the 

2-'s,  we  have 
14z=243— 5y. 

(4) 

Dividing, 

1"  ,  ^-% 

(5) 

Let 

5—5?/ 

(6) 

Multiplying, 

1472=5—5?/. 

0) 

Transposing, 

5?/=5— 1471. 

(8) 

Dividing,    . 

,         147Z 

(9) 

Reducing,  &c., 

7/=l-372+|. 

"We  might  use  the  first  value  of  ij  ;  but,  to  do  what  it  is  con- 
venient to  do  in  some  cases,  let  us  introduce  a  second  auxiliary 
quantity,  to  represent  the  fraction  in  the  2d  value  of  y. 

(10)  Let  m=^. 

0 

(11)  Multiplying,  5?n=n. 

(12)  By  substitution,  y=^ — 147?2. 

(13)  And  ,=17+!zi£=I^. 

(14)  Therefore,  z=ll-]-bm. 

The  value  of  y  requires  that  m  should  be  zero,  or  negative. 
Let  us  first  suppose  the  value  of  7)i  to  be  0. 

Then,  7/=l— 14(0)=1. 

And  z=17— 5(0)=17. 

And  2=41-1-17=23. 

Let  — 1  be  taken  for  the  value  of  ???. 

And  ?/=  l-(— 14)=  1+14=15. 

r=17+5(-l)=17—  5=12. 
.^=:41_12-15=14. 
Again,  let  — 2  be  taken  for  the  value  of  m. 

And  2/=l-14(-2)=29. 


262  A  I.  a  E  15  R  A  . 

And  ;^=:17_|_5(_2)=7. 

a:=41— 29— 7=5. 
Again,  let  — 3  be  taken  for  the  value  ofm. 

Then,  2/=l-14(-3)=43.\ 

This  value  of  y  is  more  than  the  united  values  of  x,  y,  and  z 
by  the  conditions  of  the  question. 

The  three  values  of  m  (0,  — 1,  — 2),  then,  are  the  only  ones 

•which  will  give  integral  and  positive  values  for  all  the  quantities. 

The  reason  for  using  the  second  quantity  [m)  was  to  avoid 

fractions  in  the  values  of  y  and  z.     Three  or  four  successive 

auxiliary  quantities  may  be  used  advantageously  in  some  cases. 

295.  To  find  two  square  numbers  whose  sum  shall  be  a  square. 
4.  Let  x^-\-y''=z\ 

Then  x-=z" — y^=[z-\-y){z — ?/). 

Multiplying  both  sides  by  m,  we  have  mx^=m{z-{'y){z — y). 


Assuming, 

mx=z-{-y,  and  x=7n{z — y), 

We  have, 

z-\-y=m-{z—y). 

Therefore, 

[m^-\-'V)y={m^ — Vjz=.{mr — l){7nx—y)  = 

* 

(tr- — V)mx — {jrr — \)y. 

Therefore, 

27n~y=  {m- — 1 )  mx. 

And 

2my 
x=    „  ^,. 

To  obtain  whole  numbers  without  fractions,  let  y=:m^ — 1; 
then  we  have  x=i27n,  and  z=7?i'^-\-l.  That  is,  the  general 
forins  of  the  three  numbers  will  be  a;=2??z,  y=:^in^ — 1,  and  2:= 

If  ??z=l,  we  have  x-=.  2,  ?/=  0,  and  z=2. 
772=2,        "        x=  4,  y=  3,    "    2:=5. 
m=3,       "       x=  6,  y=  8,    "    c=10. 
?7Z=4,       "        xz=^  8,  2/=.  15,    "    z—Vl. 
771=5,        "        a;=10, 7/=24,    "    2=26. 

The  pupil  will   perceive   that   the   values   of  x  and   y  may 


INDETERMINATE     ANALYSIS.  263 

represent  the  base  and  perpendicular  of  a  right-angled  triangle, 
and  z  the  hypothenuse. 

296 •  To  find  two  numbers  the  sum  of  whose  squares  is  given. 
Bj  substitution,  we  have 


6.  Find  the  values  of  x  and  y  which  will  satisfy  the  equation 
In  these  equations,  any  number  may  be  assigned  for  the  value 

of  77Z. 

If  ?7Z=1,  we  have  a;=10,  and  2/=0. 


m=2, 

(( 

x=  8, 

"    2/=6. 

7?Z=3, 

(( 

x=  6, 

"    ?/=8. 

7?Z=4, 

(i 

80 

150 

a  32  126   , 

m=S,       "        :r=:— ,    "    2/=— ,  &c. 

297.  To  find  two  square  numbers  v/hose  difi"erence  shall  be  a 
square  number. 

6.  Let  X' — y-=:z";  therefore  [x-\-y)m[x — y):=mz';  whence, 
assuming  x-\-y=mz,  and  m{x — y)=z,  we  have  x-\-y=m-[x — 2/), 
and  {7rr-{-l)y=(m^—l)x. 

Therefore,  x=l  —, — -  hj,  and  if  yz=m-—l,  then  will  x=^m^ 
-\-\,  and  z=2m. 

If  ??i=l,  we  have  x=.  2,  y=  0,  and  2r=2. 

m=2,        "       x=:  5,  y=  3,    "    z=4. 

w  =  3,        «        :f=10,  7/=   8,     "    r=6. 

??z=4.        "        x  =  VJ,  y=\b,    "    r=8,  &c. 
"We  might  assume  a  fractional  value  for  m. 


z 


264  ALGEBRA. 

298.  If  the  diflference  of  the  two  squares  be  given,  we  have 
the  following  formula  for  ascertaining  their  value ; 

m{x-\-y)=m-z,  and  m[x — y)=z. 
Whence,  2?nx={77r-\-l)z,  x={  \z,  and  y=  (  — —  J 

7.  What  values  of  x  and  y  will  satisfy  the  equation  x'^—if 
=242? 

He.  .=  (^)h  ana  ,=  (^).4, 

where  the  values  ofm  may  be  assumed  at  pleasure. 
lfm=l,  we  have  x=24:,  and  y=  0. 
m=2,       "       a;=30,    "    ?/=18. 
m=S,        "        :r=40,    "    ?/=32. 
7?2=4,        "        2:=51,    "    2/=45,  &c. 

8.  The  difference  between  the  squares  of  the  ages  of  two 
persons  at  one  period  was  45,  and  at  another  it  was  159.  Re- 
quired the  age  of  each. 

Ans.  At  the  first  period  their  ages  were  9  and  6,  and  at  the 
second  28  and  25.  Or,  at  the  first  period  they  were  23  and 
22,  and  at  the  second  80  and  79. 

EXAMPLES. 

1.  How  many  pounds  of  sugar,  at  11  cents  per  lb.,  shall  be 
mixed  with  another  kind,  at  5  cents  per  lb.,  that  the  mixture 
shall  be  worth  $2.54  ? 

Ans.  19  lbs.  with  9  lbs. ;  14  lbs.  with  20  lbs.;  9  lbs.  with 
31  lbs. ;  and  4  lbs.  with  42  lbs. 

2.  A  person  divides  65  shillings  among  15  persons,  men, 
women,  and  children.  The  share  of  a  man  is  7  shillings,  that  of 
a  woman  3  shillings,  and  that  of  a  child  2  shillings.  How  many 
persons  were  there  of  each  class  ? 

Atis.  6  men,  5  women,  and  4  children. 

3.  A  gentleman  has  two  farms,  valued  at  $2000.  The  best  is 
worth  $21  per  acre,  and  the  other  $17  per  acre.  How  many 
acres  are  there  in  each  farm  ? 


INDETERMINATE     ANALYSIS.  265 

Ans.  The  first  may  contain  92,  75,  58,  41,  24,  or  7  acres; 
and  the  second  may  contain  4,  25,  46,  67,  88,  or  109  acres. 

4.  I  purchase  wheat  at  17  shillings  and  barley  at  11  shillings 
a  bushel,  and  expend  in  all  £27  2^.  How  man}^  bushels  of  each 
do  I  purchase  ?  Atis.  6  of  wheat  and  40  of  barley ;  or 
17  of  wheat  and  23  of  barley ;  or  28  of  wheat  and  6  of  barley. 

5.  It  is  required  to  divide  100  into  two  such  parts  that  one  of 
them  may  be  divisible  by  7,  and  the  other  by  11. 

Atis.  The  only  parts  are  56  and  44. 

6.  In  how  many  ways  can  a  debt  of  $25  be  paid  with  $2  and 
$3  bills  ?  Ans.  Four  ways. 

7.  I  wish  to  mix  corn  at  70  cents  per  bushel  with  wheat  at 
$1.90  per  bushel.  How  many  bushels  of  each  must  be  taken  to 
amount  to  $9.20?         Ans.  5  bushels  of  corn,  and  3  of  wheat. 

8.  It  is  required  to  find  the  least  whole  number  which,  being 
divided  by  17,  shall  leave  a  remainder  of  7,  and,  when  divided 
by  26,  shall  leave  a  remainder  of  13.  Ans.  143. 

9.  A  person  wishes  to  purchase  20  animals  for  £20  ;  sheep  at 
31  shillings,  pigs  at  11  shillings,  and  rabbits  at  1  shilling  each. 
In  how  many  ways  can  he  do  it  ? 

Ans.  He  can  buy  12  sheep,  2  pigs,  6  rabbits;  or  11  sheep, 
5  pigs,  4  rabbits ;  or  10  sheep,  8  pigs,  2  rabbits. 
Note.  —  The  question  will  admit  of  only  these  three  answers. 

10.  It  is  required  to  find  two  numbers,  one  of  which  being 
multiplied  by  7,  and  the  other  by  13,  the  sum  of  the  products 
shall  be  equal  to  71. 

Note.  —  This  question  does  not  admit  of  an  answer  in  whole  numbers. 
No  value  can  be  given  to  the  auxiliary  unknown  quantity  (/i)j  which  will 
render  x  and  y  both  integral  and  positive. 

11.  It  is  required  to  find  two  numbers  the  sum  of  whose 
squares  shall  be  1225. 

Ans.  The  only  positive  and  integral  numbers  are  21  and  28. 

12.  The  difierence  of  the  squares  of  two  numbers  is  1521 ; 
what  are  the  numbers?  Ans.  52  and  65,  or  &c.  &c. 


266  ALGEBRA. 


SECTION    XXVIII. 

VARIATIONS,    PERMUTATIONS,    AND    COMBINATIONS. 

■      .  ^ '  '' 

Art.  299.  The  different  arrangements  that  can  be  made  of 
any  number  of  quantities,  taking  a  certain  number  at  a  time, 
are  called  Variations. 

Thus,  if  a,  b,  c,  be  taken  two  together,  the  variations  will  be 
ab,  ba,  ac,  ca,  be,  cb. 

And  if  a,  b,  c,  d,  be  taken  three  together,  their  variations  will 
be  24.     Thus, 

abc  abd  acb  acd  adb  adc 

bac  bad  bca  bed  bda  bdc 

cab  cad  cba  cbd  cda  cdb 

dab  dac  dba  dbc  dca  deb. 

If  all  the  quantities  are  taken  together,  their  variations  are 
called  Permutations. 

Thus  the  permutations  of  <2,  Z*,  c,  are  abc,  acb,  bac,  bca,  cab,  cba. 
The  permutations  of  1,  2,  3,  are  123,  132,  213,  231,  312,  321. 

The  different  collections  that  can  be  made  of  a  number  of 
things,  taking  a  certain  number  of  things  together  without  re- 
garding their  order,  are  called  Combinations.  Thus  the  com- 
binations of  a,  b,  c,  taken  two  together,  are  ab,  ac,  be. 

Each  combination  will  supply  as  many  corresponding  varia- 
tions as  the  number  of  things  it  contains  admits  of  permutations. 

VARIATIONS. 

Let  V  =  the  number  of  variations  required. 
n  =  number  of  different  things. 
7'  =  number  of  things  taken. 
The  following,  therefore,  will  be  the  formula  for  obtaining  the 
number  of  variations  of  7i  thino-s,  taken  r  to<:!fether. 


VARIATIONS.  267 

The  number  of  variations  of  n  things,  taken  r  together,  is 
n{n—\){7i—1) [7z— (r— 1)]. 

Let  a,  h,  c,  d,  &c.,  be  the  n  things;  then  the  number  of 
variations  which  can  be  made,  taking  them  singly,  is  n. 

Let  71 — 1  of  these  things,  namely,  b,  c,  d,  &c.,  be  taken 
singly ;  then  the  number  of  their  variations  is  n — 1 ;  and,  if  a 
be  placed  before  each,  we  shall  have  7i—l  variations  of  ?i  things, 
taken  two  together,  in  which  a  stands  first.  Similarly,  we  shall 
have  n — 1  such  variations  in  which  b  stands  first,  and  simi- 
larly for  all  the  n  things;  hence  there  will  be,  on  the  whole, 
{n — 1)  variations  of  n  things,  taken  two  together. 

Again,  taking  n — 1  of  these  things,  namely,  b,  c,  d,  &c.,  their 
variations,  taken  two  together,  will  be  n{n — l){n — 2)  ;  and  pro- 
ceeding as  before,  there  will  be,  in  the  whole,  {n—l){?i — 2) 
variations  of  n  things,  taken  three  together. 

Similarly,  their  variations,  taken  four  together,  will  be  n{n — 1) 
(ji — 2)(?z— 3).     Hence,  if  Fj,  Ts*  ^3>  &c-»  K?  denote  the  varia- 
tions of  7i  things,  taken  1,  2,  3,  &c.,  r,  together,  we  have 
Fi=w,  Vo=n{n-l),  V^=n{n—l){7i-2),  &c. 
V,=?i{7i-l){7i-2){7i—B) [7i-{r—l)]. 

From  the  above  we  infer  that  the  permutations  [jj)  of  n 
things  are  their  variations  taken  all  together ;  therefore,  by 
writing  7i  for  r,  we  shall  have 

p=:n{n—l){n—2) {7i—{n—2)  ){n—{7i—l)  )  = 

n{7i—l){7i-2) 2.1=1.2.3 71. 

1.  How  many  changes  can  be  rung  with  7  bells  out  of  10  ? 

V=7i{n—l){7i—2) (7^— (r— 1)). 

As  there  are  10  bells,  71=10 ;  and  as  they  are  taken  7  at  a 
time,  r=7,  and  r— 1^6;  therefore,  7i — (r — 1)=10  — 6=4. 
Hence  F-=10.9.8.7.6.5.4=604800  changes.     A7is. 

2.  How  many  words  can  be  made  with  4  letters  out  of  5  ? 

Atis.  120. 

3.  How  often  can  4  boys  change  their  places  in  a  class  of  8  so  as 
not  to  preserve  the  same  order  ?  Atis.  1680. 


268  ALGEBRA. 

PERMUTATIONS. 

300*  When  a  and  h  are  different,  their  permutations  are  ab^ 
ha  ;  but,  when  (2=^,  they  become  aa. 

Let  a  recur  "p  times ;  ^,  q  times ;  c,  r  times ;  and  P  be  the 
number    of  permutations    required.      Then,   if  all   the   a's   be 

changed  into  different  letters,  they  will  form  1.  2.  3 j9, 

permutations ;  and,  out  of  each  of  the  P  permutations,  we  should 

form  1.  2.  3 permutations.     In  like  manner,  if  all  the 

Z''s  were  changed  to  different  letters,  they  would  form  1.  2.  3 

q  permutations;  and,  therefore,  there  would  be  P.  (1.  2. 

3 jp.  1.  2.  3 q)  permutations.     Now,  when  all  the 

quantities  have  become  different,  the  number  of  permutations  is 
1.  2.  3.  4 n.  by  Art.  299. 

Therefore,  P.  (1.  2.  3 j9.  1.  2.  3 q.  1.  2.  3 r.  &c.) 

=1.  2.  3 n. 

1.  2.  3.  .  .  ?^ 


Whence,    P= 


1.  2.  3.  .  .  p.  1.  2.  3.  .  .  q,  1.  2.  3 r,  &c.* 


4.  In  how  many  ways  may  the  word  enunciation  be  written  ? 
In  this  word  there  are   11  letters,  of  which  3  are  tz's  and  2 

are  i's  ;  therefore,  ?z=ll,  _p=3,  q=l. 

„     1.  2.  3.  4.  5.  6.  7.  8.  9.  10.  11 
Hence,  P= ^   ^   o   ^^   ^ 

=  3326400  ways.     Atis. 

5.  In  how  many  ways  may  the  word  algebra  be  written  ? 

Ans.  2520. 

6.  How  many  different  numbers  can  be  made  with  the  follow- 
ing figures,  1225555  ?  Aiu.  105. 

7.  How  many  variations  may  be  made  of  the  letters  in  the 
word  zapliTiathpaaneah  1  Ans.  454053600. 

COMBINATIONS. 

301 .  The  different  collections  that  can  be  made  of  a  number 
of  things,  taking  a  certain  number  together,  without  regarding 
their  order,  are  called  their  Combbiations. 


COMBINATIONS.  269 

Thus,  the  combinations  of  «,  h,  c,  taken  two  together,  are  ah, 
ac,  he. 

Each  combination  will  supply  as  many  corresponding  varia- 
tions as  the  number  of  things  it  contains  admits  of  permu- 
tations. 

Each  combination  of  r  things  supplies  1.  2.  3 r  varia- 
tions of  r  things ;  hence,  if  C,  be  the  number  of  combinations 
of  71  things,  taking  r  together,  the  following  will  be  the  formula. 

C,  (1.  2.  3.  .  .  r)=V,=n{n-l)[n—2) {n-{r-\) ). 

iherefore,  t^=z— ^    ,^   ^ ^ • 

1.  2.  o.  .  .  .  r 

8.  Into  how  many  diflferent  triangles  may  a  decagon  be 
divided,  by  drawing  lines  from  the  angular  points  ? 

Note.  —  The  number  of  triangles  'will  be  equal  to  the  number  of  lines 
that  can  be  drawn  by  connecting  7  at  a  time  of  the  10  angles,  with  each 
angle  ;  taken  7  together, 

^     n{n—l){n-2) {n—{r—\))     10.  9.  8.  7.  6.  5.  4 

1.  2.  3 r  ~  1.  2.  3.  4.  5.  6.  7 

=  120.     Ans. 

9.  How  many  different  combinations  can  be  made  with  5 
letters  out  of  8  ?  Ans.  56. 

10.  From  a  company  of  12  persons,  it  is  proposed  to  ascertain 
how  many  parties,  of  ten  each,  can  be  selected,  and  no  two 
parties  to  be  composed  of  the  same  individuals.  How  many 
parties  can  be  selected  ?  Ans.  66. 

11.  A  company  of  soldiers  consists  of  40  men,  and  6  of  them 
are  selected  every  night  to  mount  guard ;  on  how  many  nights 
can  a  different  guard  of  6  sentinels  be  made  ?     Aiis.  3838380. 

12.  How  many  different  numbers  can  be  made  out  of  one 
unit,  two  2's,  three  3's,  and  four  4's,  supposing  all  the  figures  to 
be  in  every  number  ?  Ans.  12600. 

13.  What  is  the  total  number  of  combinations  of  16  things, 
taken  1,  2,  3,  &c.,  at  a  time  ?  Aiis.  65535. 


^  \  0,  1,  2,  3,    4,    5,    6,  indices,  or  logarithms. 

16,  32,  64,  geometrical  progression. 


0-    \l 


J 

270  ALGEBRA. 

SECTION    XXIX. 

/^^  ^-  LOGAEITHMS.^^  „><.  ^--^  .   <-cy>^ 

Art.  302.  Logarithms  are  a  series  6f  numbers  in  arith- 
metical progression,  answering  to  another  series  of  numbers  in 
geometrical  progression. 

(  0,  1,  2,  3, 

'(1248 
■c^^ejLf  '    '    '    ' 

..  (0,  1,  2,    3,    4,      5,      6,  indices,  or  logarithms. 

(  1,  3,  9,  27,  81,  243,  729,  geometrical  progression. 
0,    1,      2,        3,  4,  5,  indices,  or  log. 

10,  100,  1000,    10000,    100000,  geomet.  prog. 

From  the  above,  it  is  evident  that  the  same  indices  may  serve 
equally  for  any  geometrical  series ;  and,  consequently,  there  may 
be  an  endless  variety  of  systenis  of  logarithms  to  the  same  com- 
mon numbers,  by  only  changing  the  second  term,  2,  3,  or  10,  &c., 
of  the  geometrical  series  of  whole  numbers ;  and,  by  interpolation, 
the  whole  system  of  numbers  may  be  made  to  enter  the  geomet- 
rical series,  and  receive  their  proportional  logarithms,  whether 
integers  or  decimals. 

It  is  also  aj)parent,  from  the  nature  of  these  series,  that,  if  any 
two  indices  be  added  together,  their  sum  will  be  the  index  of 

*  The  invention  of  Logarithms  is  due  to  Lord  Napier,  Baron  of  Mer- 
chiston,  in  Scotland,  and  is  properly  considered  as  one  of  the  most  useful 
inventions  of  modern  times.  A  table  of  these  numbers  was  first  publislied 
by  the  inventor  at  Edinburgh,  in  the  year  1614,  in  a  treatise  entitled 
Canon  Mirificum  Logarithmorum ,  ^Yhich  was  eagerly  read  by  all  the 
learned  throughout  Europe.  Mr.  Henry  Briggs,  then  professor  of  geom- 
etry at  Gresham  College,  soon  after  the  discovery  went  to  visit  the  noble 
inventor  ;  after  which,  they  jointly  undertook  the  arduous  task  of  com- 
puting new  tables  on  this  subject,  and  reducing  them  to  a  more  convenient 
form  than  that  which  was  at  first  thought  of.  But,  Lord  Napier  dying 
soon  after,  the  whole  burden  fell  upon  Mr.  Briggs  ;  who,  with  prodigious 
labor  and  great  skill,  made  an  entire  canon,  according  to  the  new  form,  for 
all  numbers,  from  1  to  20000,  and  from  90000  to  101000,  to  11  places  of 
decimals,  and  published  it  in  London,  in  the  year  1G24. 


LOGARITHMS.  271 

that  number  which  is  equal  to  the  product  of  the  two  terms  in 
the  geometrical  progression  to  which  those  indices  belong.  Thus 
the  indices  2  and  3,  being  taken  together,  make  5  ;  and  the 
numbers  4  and  8,  or  the  terms  corresponding  to  those  indices, 
being  multiplied  together,  make  82,  which  is  the  number  answer- 
ing to  the  index  5. 

In  like  manner,  if  any  one  index  be  subtracted  from  another, 
the  difference  will  be  the  index  of  that  number,  which  is  equal  to< 
the  quotient  of  the  two  terms  to  which  those  indices  belong.  Thus 
the  index  G,  minus  the  index  4,  is  2 ;  and  the  terms  correspond- 
ing to  those  indices  are  64  and  16,  whose  quotient  is  4,  which 
is  the  number  answering  to  the  index  2. 

For  the  same  reason,  if  the  logarithm  of  any  number  be  mul- 
tiplied by  the  index  of  its  power,  the  product  will  be  equal  to  the 
logarithm  of  that  power.  Thus,  the  index  or  logarithm  of  4,  in 
the  above  series,  is  2 ;  and,  if  this  number  be  multiplied  by  3, 
the  product  will  be  6,  which  is  the  logarithm  of  64,  or  the  third 
power  of  4. 

And,  if  the  logarithm  of  any  number  be  divided  by  the  index 
of  its  root,  the  quotient  will  be  equal  to  the  logarithm  of  that 
root.  Thus,  the  index  or  logarithm  of  64  is  6;  and,  if  this 
number  be  divided  by  2,  the  quotient  will  be  3,  which  is  the  log- 
arithm of  8,  or  the  square  root  of  64. 

The  logarithms  most  convenient  for  practice  are  such  as  are 
adapted  to  a  geometrical  series  increasing  in  a  ten-fold  ratio,  as 
in  the  last  of  the  above  forms ;  and  are  those  which  are  to  be 
found,  at  present,  in  most  of  the  common  tables  on  this  subject. 
The  distinguishing  mark  of  this  system  of  logarithms  is,  that  the 
index  or  logarithm  of  10  is  1 ;  that  of  100,  2  ;  that  of  1000, 
3,  &c. 

In  decimals,  the  logarithm  of  .1  is  — 1,  and  that  of  .01  is 
— 2,  that  of  .001  is  — 3,  and  so  on.  The  logarithm  of  1  in 
every  system  being  0,  it  follows  that  the  logarithm  of  any  number 
between  1  and  10  must  be  0  and  some  fractional  parts,  and  that 
of  a  number  between  10  and  100  will  be  1  and  some  fractional 
part,  and  so  on  for  any  other  number  whatever.  And,  since  the 
integral  part  of  a  logarithm,  usually  called  the  Index  or  Charac- 


272  ALGEBRA. 

teristic,  is  always  thus  readily  found,  it  is  commonly  omitted  in 
the  tables ;  being  left  to  be  supplied  by  the  operator  himself,  as 
occasion  requires. 

303.  Another  definition  of  Logarithms  is,  that  the  logarithm 
is  the  index  of  that  power  of  some  other  number  whiuh  is  equal 
to  the  given  number.  So,  if  there  be  iY=r",  then  n  is  the  loga- 
rithm of  N ;  where  n  may  be  either  positive  or  negative,  or 
nothing,  and  the  root,  r,  any  number  whatever,  according  to  the 
difi"erent  systems  of  logarithms. 

When  7z  is  =  0,  then  iV  is  =  1,  whatever  the  value  of  r  is, 
which  shows  that  the  logarithm  of  1  is  always  0  in  every  system 
of  logarithms.  When  n  =  1,  then  N  =  r  ;  so  that  the  radix, 
r,  is  always  that  number  whose  logarithm  is  1,  in  every  sys- 
tem. When  the  radix  r  =  2.718281828459,  &c.,  the  indices 
n  are  the  hyperbolic,  or  Napier's  logarithm  of  numbers,  N ;  so 
that  ?i  is  always  the  hyperbolic  logarithm  of  the  number  JV,  or 
(2.718281828459)". 

804.  When  the  radix  ?•  =  10,  then  the  index  n  becomes  the 
common  or  Briggs'  logarithm  of  the  number  N ;  so  that  the 
common  logarithm  of  any  number  10"  or  iV  is  n,  the  index  of 
that  power  of  10  which  is  equal  to  the  said  number.  Thus,  100, 
being  the  second  power  of  10,  will  have  2  for  its  logarithm ;  and 
1000,  being  the  third  power  of  10,  will  have  3  for  its  logarithm. 
Hence,  also,  if  50  =  iqi-gosot^  ^^^^-^  jg  1.69897  the  common 
logarithm  of  50.  That  is,  10  has  been  raised  to  the  169897th 
power,  and  the  lOOOOOd  root  has  been  extracted,  which  is  fomid 
to  be  50,  nearly.  And,  in  general,  the  following  decuple  series 
of  terms,  namely, 

10\   10^  10-1,  10-2,  jo-3,  10-4, 
10,    1,      .1,     .01,   .001,  .0001, 
1,     0,    -1,    -2,     -3,    -4, 

for  their  logarithms,  respectively.  And  from  this  scale  of  num- 
bers and  logarithms  the  same  properties  easily  follow,  as  above 
mentioned. 

305.  To  compute  the  Logarithm  to  any  of  the  Natural  Num 
bers,  1,  2,  3,  4,  5,  &c.,  we  have  the  following 


lOS     10^ 

10^ 

or 

10000,  1000, 

100, 

have 

4,         3, 

2, 

LOGARITHMS.  273 

Rule.  Take  the  geometrical  series,  1,  10,  100,  1000,  10000, 
<f-c.,  and  apphj  it  to  the  arithmetical  series,  0,  1,  2,  3,  4,  5,  ^c, 
as  logarithms. 

Find  a  geometrical  mean  between  1  and  10,  or  between  10  and 
100,  or  any  other  two  adjacent  terms  of  the  series,  between  ivhich 
the  number  proposed  lies. 

In  like  manner,  between  the  mean  thus  found,  and  the  nearest 
extreme,  find  another  geometrical  mean;  and  so  on,  till  you 
arrive  within  the  proposed  limit  of  the  number  ivhose  number  is 
sought. 

Find,  also,  as  many  arithmetical  means  hi  the  same  as  you 
found  geometrical  ones,  and  these  loill  be  the  logarithms  ansicer- 
ing  to  the  said  geometrical  means. 

EXAMPLE. 

Calculate  tlie  logarithm  of  9. 
Here  the  proposed  number  lies  between  1  and  10. 
First,  then,  the  log.  10  is  1,  and  the  log.  of  1  is  0. 
Therefore  (l4-0)-r-2=^=.5  is  the  arithmetical  mean. 

And  (10x1)^=3.1622777,  the  geometrical  mean. 

Hence  the  log.  of  3.1622777  is  .5. 

Secondly,  the  log.  of  10  is  1,  and  the  log  of  3.1622777  is  .5. 
Therefore  (l-|-.5)-r-2  =  .75,  the  arithmetical  mean. 

And    (10x3.1622777)^=5.6234132,  the  geometrical  mean. 
Hence  the  log.  of  5.6234132  is  .75. 

Thirdly,  the  log.  of  10  is  1,  and  the  log.  of  5.6234132  is  .75. 
Therefore  (l  +  .75)~-2=.875  is  the  arithmetical  mean. 

And  (10x5.6234132)2-=7.4989422  the  geometrical  mean. 
Hence  the  log.  of  7.4989422  is  .875. 

Fourthly,  the  log.  of  10  is  1,  and  the  log.  of  7.4989422  is  .875. 
Therefore,  (1-|-.875)-t-2=.9375  is  the  arithmetical  mean. 

And  (10X7.4989422)^=8.6596431,  the  geometrical  mean. 
Hence  the  log.  of  8.6596431  is  .9375. 


274  ALGEBRA. 

Fifthly,  the  log.  of  10  is  1,  and  the  log.  of  8.6596431  is  .9375, 
Therefore,  (l-{-.9375)-i-2=.96875  is  the  arithmetical  mean. 

And  (10x8.6596431)"^=9.3057204,  the  geometrical  mean. 

Hence  the  log.  of  9.3057204  is  .96875. 

Sixthly,  the  log.  of  8.6596431  is  .9375,  and  the  log.  of 
9.3057204  is  .96875. 

Therefore,  (.9375+.96875)^2=.953125  is  the  arithmetical 
mean. 

And  (8.6596431x9.3057204)^=8.9768713,  the  geometrical 
mean. 

Hence  the  log.  of  8.9768713  is  .953125. 

By  proceeding  in  this  manner,  after  25  extractions,  it  will  be 
found  that  the  logarithm  of  8.9999998  is  .9542425,  which  may 
be  taken  for  the  logarithm  of  9,  as  it  differs  so  little  from  it,  and 
is  sufficiently  exact  for  all  practical  purposes ;  and  in  this  manner 
were  the  logarithms  of  almost  all  the  prime  numbers  at  first 
computed. 

306.  Another  method  of  computing  logarithms  is  by  the  aid 
of  a  given  decimal. 

Rule.  Let  b  he  the  nuiriher  whose  logarithm  is  required  to  be 
found,  and  a  the  nuraber  next  less  than  b,  so  that  b — a=l,  the 
logarithm  of  a  being  known  ;  and  let  s  denote  the  sum  of  the 
two  numbers,  a-[-b.     Then 

1.  Divide  the  constant  decimal  .8685889638  by  s,  and  reserve 
the  quotient ;  divide  the  reserved  quotient  by  the  square  of  s,  and 
reserve  this  quotient ;  divide  this  last  quotient,  also,  by  the  square 
of  s,  and  again  reserve  the  quotient ;  and  thus  proceed,  con- 
tinually dividing  the  last  quotient  by  the  square  of  s,  as  long  as 
division  can  be  made. 

2.  Write  these  quotients  orderly,  under  one  another,  the  first 
uppermost,  and  divide  them  respectively  by  the  odd  numbers,  1, 
3,  5,  7,  9,  ^c,  aslong  as  division  can  be  made  ;  that  is,  divide 
the  reserved  quotient  by  1,  the  second  by  3,  the  third  by  5,  the 
fourth  by  7,  a)ul  so  on. 


L  0  G  A  li  I  T  II  JI  S  . 


275 


8.  Add  all  these  last  quotients  together,  and  the  sum  will  be 
the  logarithm  of  b-j-a.  To  this  logarithm  add,  also,  the  given 
logarithm  of  the  said  next  less  number,  a  ;  the  last  sum  will  be 
the  logarithm  of  the  number  b  proposed. 


EXAMPLES. 


1.  Let  it  be  required  to  find  the  logarithm  of  the  number  2. 
Here  the  given  number  b  is  2,  and  the  next  less  number  a  is    ^CC- 
1,  whose  logarithm  is  0;  also,  the  sum  2  +  1  =  3=5,  and  its 


square  r 

=9.  Then  the  operation  will  be 

as  follows. 

3) 

8G8588964 

1) 

289529654 

289529654 

9) 

289529654 

3) 

32169962 

10 

723321 

9) 

32169962 

5) 

3574440 

714888 

9) 

8574440 

7) 

397160 

56737 

9) 

397160 

9) 

44129 

4908 

9) 

44129 

11) 

4903 

( 

446 

9) 

4903 

13) 

545 

42 

9) 

545 

15) 

61 

4 

9) 

61 

Logarithm  of  f =.301029995 
Add  logarithm  of  1=.000000000 


Logarithm  of  2z=.301029995 

2.  Compute  the  logarithm  of  the  number  3. 

Here  Z»=3,  the  next  less  number  g;=2,  and  the  sum  a-\-b: 
hz=zs,  whose  square  5^=25. 


5) 

.868588964 

1) 

.173717798 

^  .178717798 

25) 

.173717793 

3) 

6948712  ( 

2316237 

25) 

6948712 

5) 

277948 

1      55590 

25) 

277948 

T) 

11118  ( 

1588 

25) 

11118 

0) 

445  ( 

50 

25) 

445 
18 

11) 

18  ( 
Logarithm  of  ^ 

2 

=.176091260 

Logarithm  of  2  add. 

=.301029995 

Logarithm  of  8=.477121255 


276  ALGEBRA. 

307i  Because  the  sum  of  the  logarithms  of  numbers  gives  the 
logarithm  of  their  product,  and  the  difference  of  the  logarithms 
gives  the  logarithm  of  the  quotient  of  the  number,  we  may, 
therefore,  from  the  above  two  logarithms,  and  the  logarithm  of 
10,  which  is  1,  raise  a  great  many  logarithms,  as  will  appear  by 
the  following 

EXAMPLES. 

1.  To  find  the  logarithm  of  4,  we  multiply  the  logarithm  of 
2=.301030  by  2,  because  twice  2  are  4. 

Logarithm  of  2=.301030 

2 


Logarithm  of  4=. 602060 

2.  Find  the  logarithm  of  6. 

Because  2x3=6,  we  add  their  logarithms. 
Logarithm  of  2=.301030 

Logarithm  of  3=.477121 


Logarithm  of  6=.778151 

Find  the  logarithm  of  8. 

Because  2^=8,  therefore 

Logarithm  of  2=.301030 

Multiplied  by  3=  3 


Gives  logarithm  of    8=.903090 

4.  Find  the  logarithm  of  9. 
Because  3-=9,  therefore 

Logarithm  of  3=.477121 

Multii^lied  by  2=  2 

Gives  logaritlnn  of    9=. 954242 

5.  Find  the  logarithm  of  5. 
Because  -yi=5,  therefore 

From  logarithm  of         10=L000000 
Subtract  logarithm  of      2=  .301030 

Logarithm  of  5.  A7is.  .698970 


LOGARITHMS.  277 

Having  computed  bj  the  general  rule  the  logarithms  of  the 
other  prime  numbers,  7,  11,  13,  17,  19,  23,  &c.,  then,  bj  com- 
position and  division,  we  may  easily  find  as  many  logarithms  as 
we  please. 

Note. — The  index  of  every  logaritlim  is  always  one  less  than  the 
integers  to  the  given  number. 

308.  To  find  in  the  table  the  logarithm  of  any  number. 

(1.)  If  the  given  number  be  less  than  100,  or  consist  of  only 
two  figures. 

Rule.  Enter  the  first  'page  of  the  talle^  which  contains  all 
the  numbers  from  1  to  100,  and  opposite  the  given  number  will 
be  found  the  logarithm  with  the  index  prefixed. 

(2.)  If  the  given  number  be  more  than  100,  and  less  than 
1000. 

Rule.  Fi7id  the  given  number  in  the  left-hand  column  of  the 
table,  and  opposite,  in  the  next  column,  loill  be  found  the  loga- 
rithm to  which  the  index,  2,  must  be  prefixed. 

Thus,  if  the  logarithm  of  189  were  required,  we  find  this 
number  in  the  table,  and,  opposite  to  it,  we  find  the  logarithm 
.276462.     To  this  we  prefix  the  index,  2,  and  we  have  2.276462. 

(3.)  If  the  given  number  be  more  than  1000,  and  less  than 
10000. 

Rule.  Find  the  first  three  figures  of  the  given  number  in 
the  left-hand  column,  and,  opposite  to  it,  in  the  column  marked 
at  the  top  with  the  fourth  figure,  is  the  logarithm,  required.  To 
which  must  be  prefixed  the  index,  3. 

Thus,  if  the  logarithm  of  3568  were  required,  we  find  opposite 
356,  in  the  left-hand  column,  and  under  8,  found  at  the  top  of 
the  column,  .552425.  To  this  we  prefix  the  index,  3,  because 
there  are  four  figures  in  the  given  number,  thus,  3.552425. 

(4.)  If  the  given  number  be  more  than  10000. 

Rule.  Find  the  logarithm  of  the  first  four  figures  as  before, 
also  the  next  greater  logarithm ;  subtract  the  one  logarithm 
from  the  other,  as  also  their  correspondiiig  numbers,  the  one 
24 


278  ALGEBRA. 

from  the  other.  Then  say,  As  the  difference  heticeen  the  txco 
numbers  is  to  the  difference  of  their  logarithms,  so  is  the  re- 
maining part  of  the  given  number  to  the  'proportional  part  of  the 
logarithm  ;  which  part,  being  added  to  the  less  logarithm  before 
taken  out,  gives  the  whole  logarithm  nearly. 

EXAMPLES. 

1.  Find  the  logarithm  of  340926. 

The  logarithm  of  340900     is    =  .532627 

The  logarithm  of  341000     is     =  .532754 


The  differences  are  =         100  127 

Then,  as  100  :  127  :  :  26  :  33,  the  proportional  part.  This 
added  to  the  first  logarithm  (.532627+33)  gives  .532660.  To 
this  we  prefix  the  index  5,  because  the  given  number  had  six 
figures. 

(5.)  To  find  the  logarithm  of  a  number  consisting  of  an  in- 
teger and  decimal. 

Rule.  Find  the  logojrithm  of  the  decimal  part  the  same  as  if 
all  its  figures  were  integral ;  then  this,  having  prefixed  to  it  the 
proper  index,  will  give  the  logarithm  required;  remembering 
that  the  index  will  always  be  one  less  than  the  integer. 

Thus  the  logarithm  of  42.25  is  1.625827. 

(6.)  To  find  the  logarithm  of  a  proper  fraction. 

Rule.  Subtract  the  logarithm  of  the  denominator  from  the 
logarithm  of  the  numerator,  and  the  remainder  will  be  the 
logarithm  sought ;  which,  being  that  of  a  decimal  fraction,  must 
always  have  a  negative  index. 

2.  What  is  the  logarithm  of  f  |-  ? 

Logarithm  of  37  =1.568202 

Logarithm  of  94  =1.973128 


Logarithm  of  IJ  =—1.595074 

(7.)  To  find  the  logarithm  of  a  mixed  number. 


LOerARITHMS. 


279 


Rule.  Reduce  the  mixed  number  to  an  iviproper  fraction^ 
a7id  find  the  difference  of  the  logarithms  of  the  mcmerator  and 
denominator  in  the  same  manner  as  above. 

3.  What  is  the  logarithm  of  174-#  ? 


2"a 

First  17i-|=-V3'-.     Then, 
Logarithm  of  405  =2.607455 

Logarithm  of  23  =1.361728 

Logarithm  of  17^|  =1.245727 

(8.)  To  find  the  logarithm  of  any  decimal. 

Rule.  Find  the  logarithm  of  the  decimal  as  of  an  integer^ 
and  if  the  first  significant  figure  in  the  decimal  occupy  the  place 
of  tenths,  the  index  ivill  be  — 1.  Thus  the  logarithm  of  .375 
will  he  — 1.574031.  If  the  first  decimal  place  occupy  the  place 
of  hundredths,  the  index  will  be  — 2.  If  the  decimal  is  preceded 
by  two  ciphers,  the  index  vjill  be  — 3,  a?id  so  on. 

Thus  the  logarithm  of  .234  =  -1.369216 

of  .0234  =  —2.369216 

of  .00234  =  —3.369216 

of  .000234  =  —4.369216 

of  .0000234  =  -5.369216 


1.  What 

2.  What 

3.  Wliat 

4.  WHiat 

5.  What 

6.  What 

7.  What 

8.  What 

9.  What 


EXAMPLES. 

s  the  logarithm  of  1728? 
s  the  logarithm  of  23.56  ? 
s  the  logarithm  of  89632  ? 
s  the  logaritlim  of  ^^  ? 
s  the  logarithm  of  y^y  ? 
s  the  loo;arithm  of  19Tf,  ? 
s  the  logarithm  of  .3076  ? 
s  the  logarithm  of  .00016  ? 
s  the  logarithm  of  .0000006  ? 


Am.  3.237544. 

Ans.  1.372175. 

Ans.  4.952462. 
Ans.  —1.261966. 
Ans.  —2.447737. 

A71S.  1.279987. 
Ans.  —1.487986. 
Ans.  —4.204120. 
A?is.  —7.778151. 


280  ALGEBRA. 

309.  To  find  tlie  natural  number  to  any  given  logaritlim. 

This  is  to  be  found  in  the  tables  by  the  reverse  method  to  tho 
former,  by  searching  for  the  proposed  logarithm  among  those  in 
the  table,  and  taking  out  the  corresponding  number  by  inspec- 
tion, in  which  the  proper  number  of  integers  is  to  be  pointed 
oflf,  that  is,  one  more  than  the  index.  For,  in  finding  the 
number  answering  to  any  given  logarithm,  the  index  always 
shows  how  far  the  first  figure  must  be  removed  from  the  place  of 
units  to  the  left  hand,  or  integers,  when  the  index  is  aflB.rmative, 
but  the  right  hand,  or  decimals,  when  it  is  negative. 

Thus  the  number  to  the  logarithm  1.532882  is  34.11. 

And  the  number  of  the  logarithm  —1.532882  is  .3411. 

But,  if  the  logarithm  cannot  be  exactly  found  in  the  table,  we 
adopt  the  following 

Rule.  Take  out  the  next  greater  and  the  next  less^  suhtract- 
ing  one  of  these  logarithms  from  the  other,  as  also  their  natural 
numbers  the  one  from  the  other,  and  the  less  logarithm  from  the 
logarithm  proposed.  Then  say,  As  the  difference  of  the  first,  or 
tabular  logarithms,  is  to  the  difference  of  their  natural  numbers, 
so  is  the  difference  of  the  given  logarithm  and  the  least  tabular 
logarithm  to  the  corresponding  numeral  difference  ;  lohich,  being 
annexed  to  the  least  natural  number  above  taken,  gives  the 
nutural  number  sought,  corresponding  to  the  proposed  logarithm. 

EXAMPLE. 

1.  What  is  the  .natural  number  answering  to  the  given  loga- 
rithm 1.532708  ? 

Next  greater,  532754;  its  number,  341000  ;  given  log.,  532708 
Next  less,        532627;  its  number,  340900  ;  next  less,    532627 

127  100  81 

Then,  as  127  :  100  :  :  81  :  64,  nearly  the  numeral  differ- 
ence. Therefore,  340900  f  64=34.0964,  marking  off  two  in- 
tegers,  because  the  index  of  the  given  logarithm  is  1. 

Had  the  index  been  —1.532708,  its  corresponding  number 
would  have  been  .340964,  wholly  a  decimal. 


LOGARITHMS.  281 


MULTIPLICATION    OF    LOGARITHMS. 


Rule.  Take  out  the  logarithms  of  the  factors  from  the  talle^ 
then,  add  them  together,  and  their  sum  icill  he  the  logarithm  of 
the  product  required.  Then  take  out  from  the  table  the  natural 
'number  answering  to  the  sum  for  the  product  sought.  Add 
what  is  to  be  carried  from  the  decimal  part  of  the  logarithm  to 
the  affirmative  index  or  indices,  or  else  subtract  it  from  the 
negative.  Also,  adding  the  indices  together,  when  they  are  of 
the  same  kind,  both  affirmative  or  both  negative  ;  but  subtracting 
the  less  from  the  greater  when  the  one  is  affirmative  and  the 
other  iiegatice,  and  prefixing  the  sign  of  the  greater  to  the  re- 
mainder. 

examples. 

1.  Multiply  23.14  by  5.062. 

Numbers.  Logarithms. 

23.14     =     1.3643G3 

5.062     =     0.704322 


Product,  117.1343     =     2.068685 

2.  Multiply  2.581926  by  3.457291. 

Numbers.  Logaritlims. 

2.581926     =     0.411944 

3.457291     =     0.538736 


Product,  8.92647  =     0.950680 

3.  What   is   the   continued  product  of  3.902,   597.16,    and 
.0314728? 

Numbers.  Logaritlims. 

3.902  =     0.591287 

597.16  =     2.776091 

.0314728  =—2.497935 


Product,  73.335     =     1.865313 

Here  the  — 2  cancels  the  -]-2,  and  the  1  to  carry  from  the 
decimal  is  set  down. 
24^ 


282  ALGEBRA. 

5.  What  is  the  continued  product  of  3.586,  2.1046,  0.8372, 
and  0.0294  ? 

Numbers.  Logarithms. 

3.586  =      0.554610 

2.1046  =      0.323170 

0.8372  ==  —1.922829 

0.0294  =  -2.468347 


Product,  0.1857615  =  .-1.268956 

Here  the  2  to  carry  cancels  the  — 2,  and  there  remains  — 1  to 
set  down. 

DIVISION   BY   LOGARITHMS. 

Rule.  From  the  logarithm  of  the  dividend  subtract  the 
logarithm  of  the  divisor ^  and  the  number  ansiveriiig  to  the  re- 
viainder  will  be  the  quotient  required.  Change  the  sign  of  the 
index  of  the  divisor  from  affirmative  to  negative,  or  from  negative 
to  affirmative  ;  then  take  the  sum  of  the  indices,  if  they  be  of  the 
same  name,  or  their  difference,  when  of  different  signs,  with  the 
sign  of  the  greater,  for  the  index  to  the  logarithm  of  the  quotient. 
And  also,  when  1  is  borrowed  in  the  left-hand  place  of  the  decimal 
part  of  the  logarithm,  add  it  to  the  index  of  the  divisor  when 
that  index  is  affirmative,  but  subtract  it  when  negative  ;  then  let 
the  sign  of  the  index  arising  from  hence  be  changed,  and  worked 
with  as  before. 

EXAMPLES. 

1.  Divide  24163  by  4567. 

Logarithm  of  24163  =  4.383151 

Logarithm  of  4567  =  3.659631 


Quotient,  5.29078  =  0.723520 

Divide  37.149  by  523.76. 
Logarithm  of  37.149  =      1.569947 

Logarithm  of  523.76  =      2.719132 


Quotient,  .0709275  =  —2.850815 


DIVISION     BY     LOGARITnMS.  283 

3.  Divide  .06314  by  .007241. 

Logarithm  of  .06314  =  —2.800305 

Logarithm  of        .007241  =  —3.859799 


Quotient,  8.71978  =      0.940506 

Here  1  carried  from  the  decimals  to  the  — 3  makes  it  become 
-2,  which,  taken  from  the  other  — 2,  leaves  0  remainder. 
4.  Divide  .7438  by  12.9476. 

Logarithm  of  .7438  =  -1.871456 

Logarithm  of        12.9476  =      1.112189 


Quotient,  .057447  =  —2.759267 

Here  1  taken  from  the  — 1  makes  it  become  — 2  to  set  down. 

310.  To  find  the  Arithmetical  Complement  of  the  logarithm 
of  any  number. 

BuLE.     Subtract  the  logarithm  of  the  number  from  the  loga- 
rithm of  If  which  is  zero  (0). 

EXAMPLES. 

1    "What  is  the  arithmetical  complement  of  1.462398  ? 

0. 
1.462398 


-2.537602 
2.  What  is  the  arithmetical  comj)lement  of  —1.397940  ? 

0. 
—1.397940 


0.602060 
3.  What  is  the  arithmetical  complement  of  —3.678914? 

0. 
-3.678914 


2.321086 
4.  AVhat  is  the  arithmetical  complement  of  3.614582  ? 

0. 
3.614582 

-4.385418 


284  ALGEBRA. 

5.  What  is  the  arithmetical  eomplemeut  of  —4.321617  ? 

A?is.  8.678383. 

6.  What  is  the  arithmetical  complement  of  0.781562  ? 

A71S.  —1.218438. 

7.  What  is  the  arithmetical  complement  of  5.321463? 

Ans.  —6.678537. 

8.  What  is  the  arithmetical  complement  of  3.456321  ? 

Ans.  —4.543679. 

The   pupil   will   understand   the   rationale   of  this   rule,  by 
observing  that  the  product  of  a,  multiplied  by  b,  is  the  same  as 

1 

a  divided  by  — . 

Thus,  ay(^b=ab,  or  a-^-=ab. 

0 

Or,  12  multiplied  by  5  is  the  same  as  12  divided  hy  \.      •     . 

Thus,  12x5=60;  or  12-r-J-=60. 

The  same  by  logarithms. 

Logarithm  of  12,  =1.079181 

Logarithm  of  5,  =0.698970 


Logarithm  of  the  product,  60=1.778151. 

Or, 

Logarithm  of  12,  =1.079181 

Logarithm  of  i=.2=— 1.301030  Arith.  Com.  =0.698970 


Logarithm  of  the  product,  60,  =1.778151. 

Sil,  Any  number  may  be  divided  by  adding  the  arithmetical 
complement  of  the  divisor  to  the  logarithm  of  the  dividend. 
Their  sum  will  give  the  logarithm  of  the  quotient. 

9.  Divide  1728  by  12. 

Logarithm  of  1728,  =3.237544 

Logarithm  of  12=1.079181  Arith.  Com.  =—2.920819 

Ans.  144=2.158363 


INVOLUTION     BY     LOGARITHMS.  286 

10.  What  is  the  value  of  x  in  the  following  equation  ? 
>  1728x144x6 


Log. 

1728 

=     3.237544 

Log. 

144 

=     2.158362 

Log. 

6 

=     0.778151 

Log. 

36=1.556303  Arith.  Com. 

=—2.443697 

Log. 

18=1.255273     " 

=  -2.744727 

Log. 

12=1.079181     " 

=  -2.920819 

Am.  192=2.283300 
11.  "What  is  the  value  of  a;  in  the  following  equation? 
48x.75x72x.0625 


X-. 


I 


.027X120 

Log.       48  =     1.681241 

Log.      .75  =—1.875061 

Log.       72  =     1.857332 

Log.  .0625  =—2.795880 

Log.    .027=— 2.431364  Arith.  Com.  =     1.568636 

Log.     120=     2.079181       "  "      =-3.920819 

Ans.  50=1.698969 
12.  What  is  the  value  of  x  in  the  following  equation  ? 
654X320X-3691 


a:=- 


Am.  2192.28. 


87  X  9  X. 045 

13.  What  is  the  value  of  :c  in  the  following  equation  ? 
.69x7.5x32.71X.003 


x=^- 


Ans.  .000813. 


87X8908X.0008 

14.  Multiply  three  hundred  twenty-seven  ten-thousandths  by 
three  hundred  twenty-seven  thousand.  A7is.  10692.9. 

15.  What  is  the  product  of  one   thousand  and  twenty-five, 
multiplied  by  three  hundred  twenty-seven  ten-thousandths  ? 

Ans.  33.5175. 

16.  Multiply  .0716  by  1.326.  A7is.  .0949416. 

17.  Multiply  .0009  by  .009.  Aiis.  .0000081. 


286 


ALGEBRA. 


INVOLUTION   BY   LOGARITHMS. 

Rule.  Take  out  the  logarithm  of  the  given  number  from  the 
table.  Multiply  the  logarithm  thus  found  by  the  index  of  the 
poiver  proposed.  Find  the  number  answering  to  the  product, 
and  it  will  be  the  power  required. 

Note.  —  In  multiplying  a  logarithm  "with  a  negative  index  by  an  affirm- 
ative number,  the  product  will  be  negative  ;  but  that  which  is  to  be 
carried  from  the  decimal  part  of  the  logarithm  will  be  affirmative  :  and, 
therefore,  their  difference  will  be  the  index  of  the  product,  and  is  always 
to  be  made  of  the  same  kind  with  the  greater. 

EXAMPLES. 

1.  What  is  the  square  of  2.579  ? 

Logarithm  of  2.579         =         0.411451 

o 


Ans.  6.651         = 

2.  What  is  the  third  power  of  32.16  ? 

Logarithm  of  32.16         = 

Ans.  33261.9         = 

3.  Required  the  fourth  power  of  .09163. 

Logarithm  of  .09163         = 


Ans.  .000070494         = 
Here  4  times  the  negative  index  being  - 
the  difference  — 5  is  the  index  of  the  power. 


0.822902 

1.507316 
3 

4.521948 

-2.962038 
4 

-5.848152 

—  8,  and  3  to  carry, 


1 


EVOLUTION    BY   LOGARITHMS. 

Rule.  Take  the  logarithm  of  the  given  number  out  of  the 
table  ;  divide  the  logarithm  thus  found  by  the  index  of  the  root ; 
then  the  number  answering  to  the  quotient  will  be  the  root. 

When  the  index  of  the  logarithm  to  be  divided  is  negative, 
and  does  not  exactly  contain  the  divisor  without  some  remainder, 
increase  the  index  by  such  a  number  as  will  make  it  exactly 
divisible  by  the  index,  carr tying  the  units  borrovjed,  as  so  many 


EVOLUTION     BY     LOGARITHMS. 


287 


terhs,  to  the  left-hand  place  of  tJie  decimal,  and  then  divide  as  in 
ivhole  numbers. 


EXAMPLES. 

1.  What  is  the  square  root  of  365  ? 

Logarithm  of  365  = 

Ans,  19.10409  = 

2.  What  is  the  third  root  of  12340  ? 

Logarithm  of  12340         = 
Ans.  23.108         = 

8.  What  is  the  seventh  root  of  6  ? 
Logarithm  of  6         = 

Ans.  1.2917         = 

4.  Find  the  tenth  root  of  9. 

Logarithm  of  9         = 

Ans.  1.245         = 

5.  Find  the  square  root  of  .083. 

Logarithm  of     .083         = 
Ans.  .28809         = 

6.  Find  the  cube  root  of  .00059. 

Logarithm  of  .00059        = 
Ans.  .083872       = 


2.562293(2 
1.281146i. 


4.091315(3 
1.363771f. 

0.778151(7 
0.111164f 

0.954243(10 
0.095424^^^. 

-2.919078(2 
-1.459539. 


-4.770852(3 
-2.923617. 


Here  the  divisor  3,  not  being  exactly  contained  in  —4,  it  is 
auo-mented  by  2,  to  make  up  6,  in  which  the  divisor  is  con- 
tained just  2  times;  then  the  2  thus  borrowed,  being  carried 
to  the  decimal  figure  7,  makes  27 ;  which,  being  divided  by  3, 
gives  9,  &c. 

7.  What  is  the  value  of  x  in  the  following  equation  ? 


-( 


27x38xl5.61\^ 
.36X1^37" 


288 


ALGEBRA. 

Log. 

27 

=1.431364 

Log. 

38 

=  1.579784 

Log. 

15.61 

=1.193403 

Log. 

.36=- 

-1.556303  Arith.  Com. 

=0.443697 

Log. 

1.37= 

0.136721      " 

=—1.863279 

4.511527 

3 

13.534581(4 
Ans.  2419.05=3.383645 


8.  Find  the  value  of  x  in  the  following  equation. 


37    /14.21x.00208\^       .       -.qo.qq 
"=223  \ im )    •    ^-^••1^2438. 

9.  What  is  the  value  of  x  in  the  following  equation  ? 

7  /144\^  /703\^ 

10.  Find  the  value  of  x  in  the  following  equation. 

345    /872x.0065\5 

"=4i7-  {Am<md  '  ^^'  •^^^^^' 

11.  What  is  the  value  of  x  in  the  following  equation  ? 

25     /873\3   /278\^ 

12.  What  is  the  value  of  x  in  the  following  equation  ? 

17 /13.73x.0706\^      ,       -, -,000c 

18.  Find  the  value  of  x  in  the  following  equation. 
'38.47 X. 463^  ^ 


=(- 


A71S.  .887264. 


.037X576 
14.  Required  the  value  of  x  in  the  following  equation. 


,_,'J^><}]^\\     ■<■«■  128.!!. 


'-f 


COMPOUND      INTEREST.  289 

SECTION   XXX. 

COMPOUND   INTEREST. 

Art.  312.  Compound  Interest  is  interest  charged  not  only 
on  the  principal,  but  also  on  the  interest  of  preceding  years. 

Let  p  =  principal. 

r  =  rate  per  cent.,  considered  as  a  decimal,  or  hundredths. 
t  =  time  in  years. 
A  =  amount. 

Then  1  +  r  will  represent  the  amount  of  $1,  or  1£,  for  one 
year. 

And^  (\--\-r)  will  be  the  amount  of  any  principal  (7;)  for  1 
year. 

The  amount  for  two  years  will  be  p  (1-fr)  .  (l-j-7-)  = 
p[\-\-rY',  the  amount  for  3  years  will  be  p(\.-\-rY  .  (1+7") 
=p[\-\-rf\  for  4  years  it  will  be^(l+r)^  .  {l-{-r)=pil-\-rY. 

Hence,  for  any  number  of  years,  it  will  be  p[l-\-r)"',  or 
pil-^ry. 

Putting  A  for  amount,  we  have  the  following  formula  fo7 
ascertaining  the  amount  of  any  principal  at  any  rate  per  cent, 
for  any  definite  time,  at  compound  interest. 

A=p{l+rY. 

This  equation  contains  four  quantities,  A,  p,  r,  and  t ;  any 
three  of  which  being  given,  the  other  may  be  obtained.,  '  -^^V 

Thus,  we  have  the  following  ,  .u-") 

FOEMUL^. 

(1.)  A=pil+rY.  ^3  ,  .=(£)4_i. 


log.  (1+r) 

From  the  first  formula,  the  pupil  will  perceive  the  following 
25 


200  A  L  G  E  B  K  A  . 

Rule  maj  be  deduced  for  finding  the  amount  of  any  sum  at  com- 
pound interest. 

lluLE.  Add  1  to  the  ratio,  then  raise  this  sum  to  a  pmver 
lohose  ex'ponent  is  equal  to  the  time,  vmltijjly  this  pmver  hy  the 
jn'irucvpal,  and.  the  product  is  the  amount. 

By  logarithms  the  operation  is  much  facilitated,  especially 
when  the  time  is  of  much  length. 

EXAMPLES. 

1.  What  is  the  amount  of  $78.39  for  8  years,  at  6  per  cent, 
compound  interest  ? 

OPERATION   BY   THE   FIRST   FORMULA. 

^==jt7(l-l-r)'=78.39(l+.06)l 
Log.  (l-j-r)=1.06  =  0.025306 

Multiply  by  if =8,  8 


(l_f-r)'=(1.06f  ==  0.202448 

Log.  |;=78.39  =  1.894261 


74=$124.94.     A?is.  =  2.096709 

2.  What  is  the  amount  of  $144  for  6  years,  9  months,  at 
compound  interest,  at  5  per  cent.  ? 

Log.  (l+r)=1.05  =  0.021189 

Multiply  by  t,  6 


(l+,-)'=(1.05)'  =  0.127134 

Loo-.  ?,=:144  =  2.158362 


Log.  of  am.ount  for  6  years  =  2.285496 

Log.  (1.0375)  =  0.015988 

.4=8200.21.     Ans.  =  2.301484 

We  have  just  found  the  logarithm  of  the  amount  for  6  years, 
and  to  this  we  have  added  the  logarithm  of  1.0375,  it  being  the 
amount  of  $1  for  9  months,  at  5  per  cent. 


COMPOUND     INTEREST.  ""J 

3.  What  is  tbe  amount  of  $500  for  9  years,  at  G  per  ct  (j 
per  annum,  the  interest  to  be  paid  semi-annually  ? 

As  the  time,  ?,  is  to  be  calculated  in  half-years,  and  as  r  is 
considered  the  interest  of  $1   for  one  year,   therefore  2t  will 

T 

represent  the  time,  and  -  the  interest  of  $1  for  half  a  year. 

•J 

The  formula  will  therefore  be 

A=p(\  V7,)^"=500(l+.0a)i^ 

Log.  n  4-!^^  =  1.03  =  0.012837 

Multiply  by  18  half-years,  18 

Log.  ("l+^'V"'  =  0.231066 

Log.  ^=500  =  2.698970 

^=$851.21.     Ans.  =  2.930036 

4.  What   principal,    at   compound   interest,    will    amount   to 
$4000  in  10  years,  at  6  per  cent.  ? 

This  question  must  be  performed  by  the  second  formula. 

A  4000 


P 


Log.  1  06=0.025306 


(1+r)'     {IMf"' 
10 


0.253060  Arith.  Com.  =  —1.746940 
Loc.^  yl=4000  =  3.602060 


^=$2233.57.     Ans.  =  3.349000 

5.  At  what  rate  per  cent,  must  $2233.57  be,  at  compound 
interest,  to  amount  to  $4000  in  10  years  ? 

This  question  should  be  performed  by  the  third  formula. 


\p/  V'2-233.5Ty 


2PA                             /                   ALGEBllA. 

m           /ooo 

^2233.57 

/ 

.    =  3.602060 
=  3.349000 

/ 

0.253060(10 

Jog.  (l+r)=1.06 
^      -               1 

=  0.025306 

.06,  that  is,  6  per  cent.     A'tis. 

6.  In  what  time  will  $2233.57,  at  compound  interest,  at  6 
per  cent.,  amount  to  $4000  ? 

This  question  is  solved  by  the  fourth  formula. 

T         (^\       T         (^^\ 
_^'  \p)  _^'  V2233.57y_Log.  4000-log.  2233.57 
^"Log.  (l+r)~  Log.  (1+.06)    ~"  Log.  (1+.06) 

'     Log.  ^=4000  =  3.602060 

Log.  j9=2233.57  =  3.349000 


0.253060 


Log.  (l+r)=1.06  0.025306 

^,       ^  253060     ,^^ 

Thereiore  ^==————==10  years.     Ans. 

The  value  of  this  fraction  can  be  ascertained  by  logarithms. 
Thus,  Log.  253060  =  5.403223 

Log.  25306  =  4.403223 


1.000000 
t=.  10  years,  as  before. 

7.  What  will  $16  amount  to  in  30  years,  at  5  per  cent,  com- 
pound interest  ?  A71S.  $69.15. 

8.  "What  will  $2000,  at  compound  interest,  amount  to  in  11 
years,  at  8  per  cent.  ?  Ans.  $4663.31. 

9.  What  will  $27.18  amount  to  in  8  years,  3  months,  at  4  per 
cent,  compound  interest  ?  Ans.  $37.56. 


COMPOUND     I  N  T  E  11  E  S  T  .  293 

10.  What  is  the  compound  interest  of  $1728  for  8  years,  0 
months,  at  6  per  cent,  per  annum,  the  interest  to  be  paid  every 
8  months?  Ans.  $1138.74. 

11.  What  is  the  amount  of  $18.29  for  8  years,  8  months,  12 
days,  at  4  per  cent.  ?  Atis.  $25.73. 

12.  What  sum,  at  compound  interest,  will  amount  to  $800  in 
7  years,  at  5  per  cent,  compound  interest  ?         A7is.  $568.54. 

13.  What  sum  will  amount  to  $500  in  9  years,  at  6  per  cent. 
per  annum,  the  interest  to  be  paid  every  3  months  ? 

Am.  $292.54.5. 

14.  At  what  rate  per  cent,  will  $800,  at  compound  interest, 
amount  to  $1609.76  in  12  years  ?  A7is.  6  per  cent. 

15.  In  how  many  years  will  $3726  amount  to  $5007.43,  at  3 
per  cent,  compound  interest  ?  A?is.  10  years. 

-  16.  How  many  years  will  it  require  for  any  sum  to  double 
itself,  at  6  per  cent,  compound  interest  ? 

Let      2p=  the  amount. 


Then,  2p=p{l-{-ry. 

Arid       2=  (l+r)'. 

Loo-.  2 

■i.                        o 

Log.  (l-j-r)* 
Log.         2 
Lo<T.    1.06 

=0.301030 
=0.025306 

Therefore  -%%^^%\^-=zllM  years.     Ans. 

'•'  17.  How  many  years  will  it  require  any  sum  to  triple  itself, 
at  5  per  cent.  comjDound  interest?     Ans.  22  3'ears,  188  days. 

18.  In  1840,  the  number  of  inhabitants  in  the  United  States 
was  17,068,666;  in  1850,  the  number  was  23,267,498.  What 
was  the  gain  per  cent,  per  annum  ?  •       Aris.  .03146  per  cent. 

19.  At  the  same  rate  as  in  the  last  question,  in  what  year 
wiU  there  be  100,000,000  inhabitants  ?     Ans.  May  3d,  1897. 

Note.  —  This  auswer  is  on  the  presumption  that  the  census  is  taken  the 
first  day  of  May. 

25=^ 


294  ALGEBRA. 

20.  Required  the  compound  interest  upon  $155,  for  9  years, 
at  3J-  per  cent.  Ans.  56.24-f . 

21.  Required  the  amount  of  $820  for  2 J-  years,  at  4^-  per 
cent,  per  annum,  the  interest  being  paid  half-yearly. 

Ans.  $916.49+. 

22.  What  sum  at  compound  interest,  for  24-  years,  at  4^  per 
cent.,  the  interest  payable  every  six  months,  will  amount  to 
$458.25?    '■^  Ans.  $410.02. 

23.  At  what  rate  per  cent,  will  $2000,  at  compound  interest, 
amount  to  $4663.31  in  11  years  ?     "3  Ans.  8  per  cent. 

DISCOUNT   AND   PRESENT    VALUE   AT    COMPOUND   INTEREST. 

313.  Let  p  =  the  present  value. 

s  =  the  sum  due. 
t  =  the  time. 
d  =  the  discount. 

Then,  by  principles  before  explained,  we  have  the  following 

FOKMULiE. 

(1.)  j5=-.j4—  (2.)    d=s(l-—^\ 

EXAMPLES. 

1.  What  is  the  present  worth  of  $600,  due  3  years  hence,  at 
6  per  cent,  compound  interest  ?  A7is.  $503.77. 

2.  John  Smith,  Jr.,  owes  me  $312.50,  which  is  due  2  years 
hence,  at  4^  per  cent,  compound  interest.  What  sum  will  now 
discharge  the  debt  ?  Ans.  $286.16. 

3.  What  is  the  present  value  of  $1000,  due  4  years  hence,  at 

5  per  cent,  compound  interest?  Aiis.  $822.70. 

4.  What  is  the  discount  on  $3700,  due  10  years  hence,  at  5  ^^ 
per  cent,  compound  interest  ?  Ans.  $1428.51. 

5.  What  is  the  present  worth  of  $3456,  due  5  years  hence,  at 

6  per  cent,  compound  interest  ?  Ans.  $2582.52. 


DEPOSITS.  295 

■^   6.  What  is  the  discount  on  $1000,  due  four  years  hence,  at 
6  per  cent,  compound  interest  ?  Ans.  8207.91. 

7.  Rented  a  house  for  5  years,  at  $400  a  year,  the  rent  to  be 
paid  quarterly.  What  is  the  present  worth  of  this  rent,  at  8 
per  cent,  compound  interest  ?  Ans.  $1653.47. 

8.  Loaned  a  friend  $100  for  one  year,  at  2  per  cent,  per 
month,  compound  interest ;  that  is,  the  interest  is  to  be  added 
to  the  principal  each  month.  What  is  the  amount  at  the  close 
of  the  year?  Ans.  $126.82. 

9.  Which  is  the  greater  present  value,  $400  due  three  years 
hence,  at  5  per  cent,  compound  interest,  or  $500  due  4  years 
hence,  at  simple  interest  ?         Ans.  $500  is  better  by  $71.13. 

10.  What  sum  shall  I  put  into  the  Savings  Bank,  which  pays 
5  per  cent,  compound  interest,  that  shall  in  6  years  amount  to 
$1000  ?  i  ^  A.71S.  $746.21. 


SECTION   XXXI. 

DEPOSITS. 

Art.  314.  A  deposit  is  a  sum  of  money  lodged  in  the  hands 
of  some  person  or  corporation,  for  safe  keeping. 

1.  Deposited  annually  in  a  Savings  Bank,  which  pays  6  per 
cent,  compound  interest,  $144  for  20  years.  How  much  money 
shall  I  have  in  the  bank  at  the  end  of  the  20th  year  ? 

Let  a  =  the  sum  annually  deposited. 

?•  =  the  rate  of  interest. 
t  =  the  time. 
A  =  the  amount. 

By  the  rule  of  compound  interest,  the  sum  first  deposited  will 
amount  to  144(l-f-.06)"^,   or  a(l-f-r)';    for   the   second   year, 


296 


ALGEBRA 


lU{l-\-MY\  or  a{l~\-ry-^;  for  the  third  year,  144(l+.06)^ 
or  a(l+r)'-- ;  for  the  last  year,  144(l+.06)\  or  a{l-\-r)\ 

315.  We  have  now  a  regular  series  in  Geometrical  Progres- 
sion, where  the  extremes  are  a{l4-ry  and  a(i-\-ry,  the  ratio 
l-j-r,  to  find  the  sum  of  the  series. 

Hence,  by  Art.  276,  we  have  the  following  formula  for  obtain- 
ing the  amount  of  the  deposits. 

4l+r)[(l+r)'-l] 


Xl.= 

r 

OPERATION   BY   LOGARITHMS. 

Log.  (l+r)=1.06 

=0.025306 

Multiply  by  t=20 

3.207 

=            20 

Log.  (l+r)'=20 

=0.506120 

Subtract 

1 

Log. 

2.207 

=0.343802 

Log.  (l+r)=1.06 

=0.025306 

Log.  a=14:4: 

=2.158362 

Log.  r=.06=— 2.< 

r78151  Arith. 

Com. 

=1.221849 

Ans.  $5614.60=3.749319 

2.  A  gentleman  has  a  daughter,  who  is  10  years  old ;  and  he 
wishes  to  give  her,  as  soon  as  her  age  shall  be  21  years,  $2000. 
What  sum  must  he  deposit  annually  in  a  bank,  which  pays  5 
per  cent,  compound  interest,  to  be  able  to  accomplish  it  ? 

816,  The  question  given  above  may  be  solved  by  the  folio w- 
inor  formula,  which  is  obtained  from  the  last  by  transposition, 


in 
&C. 


Ar 


2000X.05 


a= 


Log.  2000 
Log.     .05 


(l-fr)[(l+r)'-l].     (1.05).[(l+.05)"-l]- 

OPERATION    BY   LOGARITHMS. 

3.301030 
-2.698970 

From     2.000000 


DEPOSITS.  297 


Log.  1.05=0.021189 
11 


1.71=0.233079 
1 


Los.    .71  =-1.851258 


'& 


Loir.  1.05  =     0.021189 


Take     -1.872447 


Ans.  $134.14.     =     2.127553 

3.  A  gentleman,  when  his  daughter  was  10  years  old,  de- 
posited for  her,  annually,  $134.14  in  a  bank,  which  paid  5  per 
cent,  compound  interest.  This  sum  remained  until  the  time  of 
her  marriage ;  the  amount  then  was  $2000.  What  was  then 
her  age  ? 

317.  The  formula  for  the  operation  of  the  above  question  is 
obtained  from  the  former  by  transposition,  &c. 

^       /     ^r    \  ,  ,  ,       /     2000X.05     \  ,  ^ 

^=I-^g\j(l+^j+^    =^^g\l34l4(r+:05)j+^ 
Log.      (l+r)  Log.  (1.05) 

Log.  2000  =3.301030 

.05  =—2.698970 


From         2.000000 


Log.  1.05  =0.021189 

Log.  134.14  =2.127553 

Take         2.148742 


0.71  =—1.851258 

1 


1.71  =0.232996 

.232996-^.021189  =  11  years,  nearly. 
104-11=21  years.     Ans, 


Ii98  ALGEBKA. 

4.  A  certain  town  in  the  United  States,  at  the  beginning  of 

,  1840,  had  1000  inhabitants.  There  has  been  an  emigration  to 
this  town  each  successive  year,  on  the  1st  of  January,  of  1000 
additional  inhabitants.  Now,  supposing  the  population  each 
year  to  gain  3  per  cent.,  how  many  inhabitants  would  there  be 
in  this  town  at  the  end  of  10  years  ?  Aris.  11,807. 

5.  A  gentleman,  at  the  time  of  his  marriage,  deposited  in  a 
I    savings'  bank,  for  the  use  of  his  wife,  the  sum  of  $150.     This  he 

continued  to  do  for  every  six  months  until  she  was  fifty  years 
old.  Now,  if  the  bank  pay  a  semi-annual  dividend  of  2  per 
cent,  compound  interest,  and  the  gentleman's  wife  at  the  time 
of  her  marriage  was  25  years  old,  what  is  the  amount  of  the 
deposits?  Ans.  $12,939.97. 

■^  6.  If  a  man  deposits  annually  in  a  bank  $47,  in  how  long 
time  will  it  amount  to  $400,  at  6  per  cent,  compound  interest  ? 

Ans.  6  years,  273  days. 

"^  7.  A  gentleman  has  a  son  who  is  15  years  old,  and  a  daughter 
who  is  10  years  old.  He  intends  that  each  of  them,  at  the  age 
of  21,  shall  have  $5000  in  a  savings'  bank,  which  pays  an 
annual  dividend  of  4 J-  per  cent.  What  sum  shall  he  deposit 
annually  for  each  ? 

Ans.  $712.48  for  the  son,  $345.71  for  the  daughter. 

8.  Deposited  annually,  in  a  bank  which  pays  4  per  cent.,  com- 
pound interest,  a  certain  sum,  which  in  10  years  amounted  to 
$300.     What  was  the  annual  deposit  ?  Ans.  $24.02,8. 

9.  A  certain  young  lady  deposited  $10  in  a  savings'  bank, 
and  this  she  continued  every  three  months.  Now,  if  the  bank 
pays  Ij-  per  cent,  compound  interest  at  the  end  of  each  quarter, 
what  will  be  the  amount  of  her  deposits  in  10  years  ? 

Ans.  $550.81. 


10.  Now,  if  the  lady  in  the  last  question  had  deposited  $40 
annually  at  the  commencement  of  each  year,  and  had  received  G 
per  cent,  compound  interest,  would  her  deposits  at  the  end  of 
10  years  have  been  more  or  less  than  before  ? 

Am.  $8.02  more. 


EXPONENTIAL    OR   TRANSCENDENTAL   EQUATIONS.  299 

SECTION   XXXII. 

EXPONENTIAL    OR   TRANSCENDENTAL   EQUATIONS. 

Art.  318.  To  what  power  must  7  be  raised  to  amount  to 
2401? 

Let  X  be  the  power. 
Then  7"= 2401. 

The  second  power  of  7  is  found  by  multiplying  the  logarithm 
of  7  by  2 ;  and  the  fifth  power  of  7  is  found  by  multiplying  the 
logarithm  of  7  by  5,  see  Art.  300 ;  therefore  the  x\\s.  power  of 
7  is  found  by  multiplying  the  logarithm  of  7  by  x. 

We  have,  therefore,  the  following  equation,  the  logarithm  of  7 
being  0.845098,  and  the  loo;arithm  of  2401=3.380392. 

:6-x0.845098=3.380392. 

m,       ^  3.380392     ,  , 

Ihereiore,  .7:=^  ^, .. .  ,,^,=r:4th  power.     Ans. 

The  value  of  x  is  obtained  by  dividing  the  logarithm  of  the 
numerator  by  the  logarithm  of  the  denominator. 

The  value  of  the  logarithms  may  also  be  obtained  by  sub- 
tractino-  the  loo;arithm  of  the  denominator  from  the  loo;arithm  of 
the  numerator,  and  finding  the  value  of  the  remainder.     Thus. 

Log.  3.380392  z=     0.528967 

Log.  0.845098  =—1.926907 


Ans.  4th  power,  as  before,       =     0.602060 

S19.  If  the  form  of  the  equation  be  x^^z^a^  the  value  of  x 
may  be  found  by  the  following 

KuLE.  Firsts  find  by  trial  two  numbers  as  near  the  true 
value  of  X  as  possible^  and  siibstitute  them  for  x  separately. 
Then  say,  As  the  difference  of  the  results  is  to  the  difference  of 
the  two  assumed  numbers.,  so  is  the  difference  of  the  true  result, 
and  either  of  the  former,  to  the  difference  of  the  true  number  and 
the  supposed  one  behnging  to  the  result  last  used.     Add  this  dif- 


300  ALGEBRA. 

ference  to  the  supposed  number.,  or  subtract  from  it,  according  as 
it  may  be  either  too  little  or  too  great,  and  it  will  give  the  true 
value  nearly. 

EXAMPLES. 

1.  What  is  the  value  of:?:  in  the  following  equation,  a;*=100? 

Here  ^Xlog-  cc=\og.  100=2. 

We  find  the  value  of  x,  upon  trial,  to  be  between  3  and  4. 

Log.        3=       0.477121 

Log.        4=       0.602060 

Log.  3x3=0.477121x3  =1.431363 

Log.  4x4=0.602060x4  =2.408240 


Difi'erence  of  results  =0.976877 

2.000000 
1.431363 


Difference  from  the  true  result        =  .568637 

Therefore,         .976877  :  1  :  :  .568637  :  .582 
34-.582=3.582=:z.-  nearly. 

This  value  of  x  is  found,  on  trial,  to  be  too  small,  and  3.6  is 
found  to  be  too  great ;  therefore,  by  substituting  each  of  these, 
we  have 

Log.  3.582  =0.554126 

Log.  3.6  =0.556303 

Log.  3.582x3.582=0.554126x3.582=1.984879 

Log.  3.6     X36      =0.556303x3.6     =2.002690 


0.017811 
3.6— 3.582=.018 ;  2.000000—1.984879=0.015121. 

Then  .017811  :  .018  :  :  0.015121  :  .0152. 

Therefore,     .0152+3.582=3.5972,  very  nearly. 

2.  Griven  a:''^10  to  find  x. 
First,  let  x=2.b. 
Then  log.  2.5  =0.397940. 


EXPONENTIAL    OR   TRANSCENDENTAL    EQUATIONS.  301 

And  0.397940X2.5  =  .994850. 

Secondly,  let  x=2.Q. 

Then  log.  2.G  =0.414973. 

And  0.414973X2.6  =1.078929. 

1.078929— .994850  =  .084079. 

1.-.994850=.005150;  2.6-2.5=.l. 
Then  .084079  :  .1  :  :  .005150  :  .006. 
2.5+.006=2.506,  nearly. 

3.  Required  the  value  of  x  in  the  following  equation  : 

a;^=256.  Ans.  x=4. 

4.  Given  x''=b  to  find  the  value  o?  x.  Aiis.  a;=2.129. 

5.  Required  the  value  of  a;  in  the  following  equation : 

7^=343.  Ans.  a;=3. 

6.  Find  the  value  of  x  in  the  following  equation :  a:'^=3125. 

Ans.  x=b. 

320t  This  rule  will  apply  to  solving  questions  in  geometrical 
progression,  when  we  wish  to  obtain  the  number  of  terms. 

EXAMPLES. 

7.  If  the  first  term  is  5,  the  last  term  405,  and  the  ratio  3, 
what  is  the  number  of  terms  ? 

In  Art.  274,  we  find  X=«r""^  and  this  equation,  by  trans- 
position, &c.,  is 

/  ^°'  \  a)  .  T      Log.  L— Log,  a 

I       n=i—r- f-i= Y r-*^- 

Log.  r  Log.  r 

OPERATION. 

Log.  405  =     2.607455 

Log.      5  =     0.698970 


1.908485 
Log.  3=0.477121 

Log.  1.908485  =     0.280688 

Log.  0.477121  =-0.678628 


4     =       .602060 
4-|-l=5,  the  number  of  terms.     Ans. 
26 


302  ALGEBRA. 

8.  If  the  first  term  is  4,  the  ratio  3,  and  the  sum  of  the  series 
484,  what  is  the  number  of  terms  ? 
In  Art.  278,  we  find 

^     ar"—a        alf'—l) 
S= -,  or  -j ---. 

Therefore,  by  transposition,  we  have 
Log.  [a-f-(r— I)S]-Log.  a_Log.  [4-[- (8 -1)484] -Log.  4 
Log.  r  Log.  3 

=  [4-f(3— 1)484]— 4=972— 4. 
Log.  972  ^  ==     2.987666 

Log.     4  *  =     0.602060 


2.385606 


Log.  3=.477121 

Log.  2.385606  =  0.377598 

Log.  .477121  =-0.678628 


=     0.698970 
Ans.  5,  the  number  of  terms. 

9.  How  long  must  $78.39  be  at  compound  interest,  at  6  per 
cent.,  to  amount  to  $124.94  ?  Ans.  8  years. 

10.  January  1,  1840,  lent  my  friend  John  Brown  $2000,  at 
8  per  cent,  compound  interest,  and  he  agreed  to  pay  me  in  5 
years ;  but,  owing  to  certain  circumstances,  he  could  not  pay 
until  the  amount  of  the  note  was  $4663.31.  AVhen  was  the  note 
paid  ?  Ans.  January  1,  1851. 

11.  How  long  will  it  require  $800,  at  6  per  cent,  compound 
interest,  to  amount  to  $1609.76  ?  A.ns.  12  years. 

12.  Loaned  $2000,  at  compound  interest,  for  11  years,  and 
received,  interest  and  principal,  $4663.31.  At  what  rate  per 
cent,  was  the  money  lent  ?  Ans.  8  per  cent. 

13.  A  gentleman  agreed  with  another  to.  board  him  for  a 
certain  number  of  days,  on  the  following  terms  :  he  was  to  pay 
3  cents  for  the  first  day's  board,  9  cents  for  the  second  day,  27 
cents  for  the  third  day,  and  so  on  in  this  ratio.  The  amount  of 
the  gentleman's  bill  was  $295.23.  IIow  many  days  was  the 
gentleman  boarded  ?  Ans.  9  days. 


ANNUITIES.  30»^ 

SECTION  XXXIII. 

ANNUITIES. 

Art.  321  •  Annuity  is  a  term  used  for  any  periodical  income 
arising  from  money  lent,  or  from  tenements,  land,  salaries, 
jiensions,  <fec.,  payable  from  time  to  time,  but  generally  by 
annual  payments. 

322,  Annuities  are  divided  into  those  that  are  in  Possession, 
and  those  that  are  in  Reversion  ;  the  former  meaning  such  as 
have  commenced,  and  the  latter  such  as  will  not  begin  till  some 
particular  event  has  happened,  or  till  after  some  certain  time 
has  elapsed. 

323 1  When  an  annuity  is  forborne  for  some  years,  or  the 
payment  is  not  made  for  that  time,  the  annuity  is  said  to  be  in 
arrears. 

324 1  An  annuity  may  also  be  for  a  certain  number  of 
years ;  or  it  may  be  without  any  limit,  and  then  it  is  called  a 
perpetuity. 

325.  The  amount  of  an  annuity,  forborne  for  any  number  of 
years,  is  the  sum  arising  from  the  addition  of  all  the  annuities 
for  that  number  of  years,  together  with  the  interest  due  upon 
each  after  it  became  due. 

826.  The  present  ivorth,  or  value  of  an  annuity,  is  the  price 
or  sum  which  ought  to  be  given  for  it  at  the  present  time. 

EXAMPLES. 

1.  A  man  is  desirous  to  bequeath  his  son  a  certain  sum  or 
money,  which  shall  be  deposited  in  an  annuity  office,  that  pays 
ti  per  cent.,  that  his  son  may  receive,  at  the  close  of  each  year, 
$100  for  the  term  of  12  years,  at  which  time  the  principal  and 
interest  shall  be  exhausted.     What  is  the  sum  bequeathed  ? 

Let  ^1  =  the  sum  put  at  interest. 

a  =  the  sum  taken  out  annually. 
r  =  the  rate  per  cent. 
t  =  the  time. 


304  ALGEBRA. 

327.  The  amount  of  the  sum,  a^  takeu  out  at  the  close  of 
the  first  year,  would  be,  at  the  end  of  the  time,  100(l-|-.06)^^, 
or  a(\.-\-r\~^ ',  that  taken  out  at  the  close  of  the  second  year 
would  amount  to  100  (l+.06)i'',  or  ^(l-fr)'-^;  that  taken 
out  at  the  end  of  the  third  year  would  be  100(l-f-.06)^  or 
«(l-|-r)'~^;  that  taken  out  at  the  end  of  the  12th  year  would  be 
only  a,  or  $100  without  interest. 

Thus,  we  have  a  regular  series  in  Geometrical  Progression, 
where  we  have  the  extremes,  a  and  <2(1 -]-?*) '~^,  and  the  ratio 
(1  +  r),  given  to  find  the  sum  of  the  series. 

Therefore,  by  Art.  277,  we  find  the  sum  of  the  series  to  be 

r  r  .         r 

of  all  the  sums  deposited.     This,  by  the  hypothesis,  must  be 
equal  to  A{\-\-r)\ 

Therefore,  ^(l+r)'=:ffil±^^^^l 

By  division.  A^=.         .. — ^=  sum  put  at  interest. 

We,  therefore,  have  the  first  of  these  formulae  for  finding  the 
amount  of  the  sums  drawn  out  annually,  or  at  stated  periods ; 
and  the  last  formula  for  ascertaining  what  sum  must  be  de- 
posited, or  put  at  interest. 

^_«[(l+r)'-l]_100[(1.06r-l] 


r(l+r)'  .06(14-.06) 


OPERATION   BY  LOGARITHMS. 


12 


Log.  l+r=1.06=0.025306 

12 


2.0122    =0.303672 
1 


Log.  1.0122  =0.005266 

Log.  100  =2.000000 


From  2.005266 


ANNUITIES.  305 

Log.  (l+r)'=(1.06)^  =0.303672 

Log.  r=    .06  =-2.778151 

Take     —1.081823 


$838.38.     Ans.     =2.923443 

2.  A  gentleman  deposited,  in  an  annuity  office,  $2000.  How 
mucli  can  he  receive  annually,  if  the  annuity  continue  15  years, 
at  5  per  cent,  compound  interest  ? 

By  transposition,  &c.,  of  the  last  formula,  we  obtain  the  fol- 
lowing for  ascertaining  the  value  of  the  annuity,  a. 

^         _Ar{l+ry  _2000X.05(1.05)^^ 

^— (l_fr)'— r  *      ~"      (1.05)1^—1      * 
Log.  14-r=1.05=  0.021189 

15 


(l-|_r)'=2.0789=  0.317835 

1. 


Log.         1.0789=0.032981  Arith.  Com.  =—1.967019 

Log.  (^)=2000  =     3.301030 

Log.    (r)=.05  =-2.698970 

Log.  (l+r)'=(1.05)^*  =     0.317835 

a=S192.68.     A71S.  =     2.284854 

In  the  operation  of  the  above  question,  we  find  it  more  con- 
venient to  commence  with  the  denominator  of  the  formula. 

3.  A  gentleman  deposited  in  an  annuity  office,  which  pays  5 
per  cent,  compound  interest,  $8000;  in  how  many  years  will 
this  sum  be  exhausted,  if  he  draw  out,  annually,  $850  ? 

328.  From  the  equation,  A=  ^^    .,  \ — rr-^,    we    obtain,   by 
^  ?-(l-j-r)' 

transposition,  &c.. 


t= 


^'^(■^r)        ^'^{m-isLx.ob)) 


'Log.  (1+r)   ■  Log.  (1.05) 

26=* 


306  ALGEBRA. 

Log.  (^)=850  =2.929419 

^r=8000x.05=400 
Log.  (850— 400)  =450  =2.653213 


0.276206 
Log.  (l+r)=1.05  =0.021189 

276206 
Therefore,  =13.035=13  years,  12  days.     Ans. 

329.  But  the  same  result  will  be  obtained  by  subtracting 
the  logarithm  of  the  denominator  from  the  logarithm  of  the 
numerator,  and  finding  the  number  corresponding  with  the  re- 
mainder.    Thus, 

Log.  276206  =5.441233 

Log.    21189  =4.326110 


Ans.  13.035=13  years,  12  days,  =1.115123 

4.  John  Smith,  believing  he  shall  live  20  years,  has  purchased 
an  annuity,  which  affords  him  $500  each  year.  What  sum  has 
he  deposited  in  the  annuity  office,  which  pays  for  deposits  5  per 
cent,  compound  interest  ?  The  principal  and  interest  are  to  be 
exhausted  at  the  close  of  the  20th  year.  Ans.  $6230.81. 

5.  If  John  Smith  die  at  the  end  of  10  years,  what  sum  will 
remain  in  the  office?.  An^.  $3850.27. 

6.  Or,  if  the  office  have  agreed,  for  his  deposit,  to  give  him, 
at  the  close  of  each  year,  $500,  and  if  Smith  should  live  30 
years,  what  will  the  office  lose  ?  Ans.  $6289. 

•  7.  A  gentleman  bequeathed  to  his  wife  $1728,  which  she 
deposited  in  an  office  which  pays  4  per  cent,  compound  interest. 
How  large  a  sum  shall  she  receive,  annually,  from  the  office,  that 
the  annuity  may  continue  10  years  ?  Ans.  $213.09. 

8.  A  certain  Savings  Bank  will  pay  1^  per  cent,  compound 
interest,  semi-annually.  If  I  deposit  in  this  bank  $4000,  and 
take  from  it,  at  the  end  of  every  six  months,  $500,  in  what  time 
shall  1  have  withdrawn  all  my  money  from  the  bank  ? 

Ans.  4  years,  106  days. 


INVOLUTION     0¥     BINOMIALS.  307 

9.  What  sum  shall  I  deposit  in  an  annuity  office,  that  I  may 
draw  on  it  every  3  months  for  $90  ?  The  bank  pays  on  deposits 
1  per  cent,  each  quarter  of  the  year,  and  I  wish  to  continue 
drawing  on  the  bank  for  10  years.  Atis.  $2954.84. 

-»  '    r      ,     ^       I  '■■-     'i    -"'      ■      \ 


SECTION    XXXIV. 

INVOLUTION    OF   BINOMIALS. 

Art.  330.  A  binomial  or  residual  quantity  may  be  raised 
to  any  power,  without  the  trouble  of  continual  involution,  by  the 
following 

IluLE.    1.   To  find  the  terms  without  the  coefficients. 

The  index  of  the  firsts  or  leading  quantity,  begins  ivith  the 
index  of  the  given  power  ;  and,  in  the  succeeding  terms,  decreases 
continually  hy  1,  in  every  term,  to  the  last ;  and  in  the  second, 
or  following  quantity,  the  indices  of  the  terms  are  0,  1,  2,  3,  4, 
SfC,  increasing  by  1.  That  is,  the  first  term  will  contain  only 
the  first  part  of  the  root,  ivith  the  same  index  as  the  required 
power.  The  last  term  of  the  series  vMl  contain  only  the  second 
part  of  the  given  root,  raised  to  the  intended  power  ;  but  all  the 
other  intermediate  terms  will  contain  the  product  of  sow.e  powers 
of  both  members  of  the  root,  that  the  powers  or  indices  of  the  first 
or  leading  Tnember  will  always  decrease  by  1,  ivhile  those  of  the 
seco7id  member  will  increase  by  1. 

2.   To  find  the  coefiicients. 

The  first  coefficient  is  always  1,  and  the  second  is  the  some  as 
the  index  of  the  required  power  ;  to  obtain  the  third  coefficient, 
multiply  that  of  the  second  term  by  the  index  of  the  leading  letter 
in  the  same  term,  and  divide  the  product  by  2,  and  so  on  ;  that 
is,  multiply  the  coefficient  of  the  term  last  found  by  the  index  of 
the  leading  quantity  in  that  term,  and  divide  the  product  by  the 
number  of  terms  to  that  place,  and  it  will  give  the  coefficient  of 
the  term  next  folloioing.  In  this  manner  all  the  coefficients  will 
be  obtained. 


308  A  L  G  i;  B  11  A . 

NoTK  1.  —  The  whole  number  of  terms  will  be  one  more  than  the  index 
of  the  given  power  ;  and,  when  both  terms  of  the  root  are  -[-,  all  the  terms 
of  the  power  will  be  -|-  ;  but,  if  the  second  term  be  — ,  all  the  odd  terms 
will  be  -}-,  and  all  the  even  terms  — ,  which  causes  the  terms  to  be  -j- 
and  —  alternately. 

Note  2.  —  The  sum  of  the  two  indices  in  each  term  is  always  the  same 
number,  that  is,  the  index  of  the  required  power  ;  and,  reckoning  from 
the  middle  of  the  series,  both  ways,  or  towards  the  right  and  left,  the 
indices  of  the  two  terms  are  the  same  figures  at  equal  distances,  but 
mutually  changed  places.  Also,  the  coefficients  are  the  same  numbers  at 
equal  distances  from  the  middle  of  the  series  towards  the  right  and  left ; 
so,  by  whatever  numbers  they  increase  to  the  middle,  by  the  same,  in  the 
reverse  order,  they  decrease  to  the  end. 

EXAMPLES. 

1.  Let  a-\-x  be  involved  to  the  5th  power. 

The  terms  without  the  coefficients,  by  the  first  rule,  will  be 

a'^,  a^x,  a^x",  a'x^,  ax*,  x^, 
The  coefficients  by  the  second  rule  will  be 

5X4  10X3   10X2  5X1  _ 
'    '     2    '      3     '      4     '     5    '"" 
1,  5,  10,  10,  5,  1. 
Therefore,  the  fifth  power  with  the  coefficients  is 
a,^,  ba^x,  10^"^;-,  10a-:t'^,  bax^,  xP. 

2.  Involve  a — x  to  the  sixth  power. 

jhis.  The  terras  with  the  coefficients  will  be, 

({^—Qa'x-\-lba'x"-md'x^-\-lba'x'—QaxP-]-x\ 

3.  Required  the  tenth  power  of  <z4-'^'- 
a}^-\-l0d^x-\-^ba^x'-\-120dJx^-\-1\M'x'-\-2b2a'x' 


^^^^'    '  +210aV-fl20aV+45fl"z«+10aa;«+a;i'^. 

4.  liaise  x-\-y  to  the  seventh  power. 

Am.  a;^+7a:«2/+21ry+352-V+B5.ry+21a;V+7a:/+2/'- 

5.  What  is  the  ninth  power  o^  a—b? 

Ma'b'-\-^ab''-Ij\ 


INVOLUTION     or     BINOMIALS. 


309 


The  coefficients  of  the  first  twelve  powers  will  be  found  in  the 
following 

TABLE. 

First  power,  1,  1 

Second       "  1,  2,  1 

Third         "  1,  3,  3,  1 

Fourth       "  1,  4,  6,  4,  1 

Fifth         "  1,  5,  10,  10,  5,  1 

Sixth         "  1,  G,  15,  20,  15,  C,  1 

Seventh    "  1,  7,  21,  35,  35,  21,  7,  1 

Eighth      "  1,  8,  28,  56,  70,  56,  28,  8,  1 

Ninth       "  1,  9,  36,  84,  126,  126,  84,  36,  9,  1 

Tenth       "  1,  10,  45,  120,  210,  252,  210,  120,  45,  10,  1 

Eleventh  "  1,  11,  55,  165,  330,  462,  462,  330,  165,  55,  11,  1 

Twelfth     "  1,  12,  66,  220,  495,  792,  924,  792,  495,  220,  66,  12,  1. 

By  examining  the  preceding  table,  we  readily  perceive  the 
law  by  which  the  coefficients  are  obtained. 

If  wc  wish  to  obtain  the  coefficients  of  the  6th  power,  we 
add  together  the  coefficients  of  the  5th  power,  two  and  two. 

Thus,  1+5=6;  5+10=15;  10+10=20;  10+5=15; 
5  +  1=6.  By  this  process  we  obtain  all  the  coefficients  of  the 
6th  power,  except  the  first  and  last,  which  are  always  1  in  every 
power. 

To  obtain  the  coefficients  of  the  10th  power,  we  add  those  of 
the  9th.     Thus, 

1+9=10;  9+36=45;  36+84=120;  84+126=210;  126 
+126=252;  126+84=210;  84+36=120;  36+9=45;  9 
+  1=10. 

Wc  therefore  find  the  coefficients  to  be, 

1,  10,  45,  120,  210,  252,  210,  120,  45,  10,  1. 

6.  Raise  a+43  to  the  third  power. 
Let  71=4:5. 

Then  a-\-n=:a-{-4:b. 

The  third  power  of  a+7^,  by  Art.  330,  = 


310  ALGEBRA. 

Substituting  43  for  n,  we  have 

a^-\-  Za-{U)  +  3^(43)24-  (43)^= 
a'-\-\2a-b-{-A.'iab'-\-Ub\     Ans. 

7.  What  is  the  third  power  of  a-\-b-{-c  ? 
Let  n=b-\-c. 
Then                           a-\-n=a-{-b-}-c. 

The  third  power  of      a-\-7i=a'-\-oa'^n-\-'6a?i^-\-?t\ 
Substituting  the  values  of  7t,  we  have 

a"-]-na-{b-{-c)-]-Sa{b-]-cY-{-{b+cf= 

(  aP-{-Sa-b-]-Sa-c-{-SaP-\-Qabc-{-Bac^ 
^^^'    I      j^b^j^Sb'c-Jr^bc^'+c\ 

8.  What  is  the  3d  power  of  a+b^c-{-d? 

Let  x=a-\-b,  and  y=c-\-d.  « 

Then  (^x-\-yY={a-{-b-\-c-{-d)\ 

And  ^x-\-yY={x^-{-nxhj+  Sxy'+fh 

Substituting  these  values  of  x  and  y,  we  have 

(^a+bf+^{a+b)%c+d)-{-^ai-b){c+dr-i-{cJrdY= 

ci'+'da-b~{-Sab^-i-P-\-{Sa^-\-6ab+Sb-){c-\-d)+  {3a-\'^b){c''  -{-2cd 
+d')-[-{c''+3rd-\-Scd'-j-d')= 

a'-{-Sa'b-\-oaF^+b^-i-^a-c^Qabc+db-c-i-SaM-^Qabd-^W-dJr 
3ar-^Qacd-\-Sad'-j-Sbc''-^6cbd-^Sbd'Jl-c^Jr^c''d-\-dcd--^dK  A?is. 

9.  What  is  the  3d  power  of  2a—b-\-r  ?  Ans. 

10.  What  is  the  5th  power  of  4(2 — bb?  A71S. 

11.  What  is  the  6th  power  of  3^-— SZ*^  ?  A71S. 

12.  What  is  the  4th  power  of  7n-\-7i — p  ?  A71S. 

13.  What  is  the  8th  power  of  m^-|-?i^?  A71S. 

14.  What  is  the  7th  power  of  1-j-^'  ?  A7is. 

15.  What  is  the  2d  p'ower  of  a-{-b-{-c-{-d-\-e-\-f  ?  Ans. 

16.  What  is  the  10th  power  of  a^-\-b^  ?  A7U 

17.  What  is  the  7zth  power  of  a-{-b?  Ans. 

18.  AVhat  is  the  6tli  power  of  a — b-\-c?  A71S. 

19.  What  is  the  4th  power  of  a^ — x  ?  A71S. 

20.  What  is  the  3d  power  of  2a?— U^  ?  Aois. 


BINO.AIIAL     TIIEOIIEM.  311 


SECTION  XXXV. 

BINOMIAL   THEOPvEM. 

Art.  S31.  The  Binomial  Theorem  is  a  general  algebraical  ex- 
pression or  formula,  by  ^Yhich  any  po^er  or  root  of  a  given 
quantity,  consisting  of  two  terms,  is  expanded  into  a  series  ;  the 
form  of  which,  as  it  was  first  proposed  by  Sir  Isaac  Newton, 
being  as  follows : 


.:>n 
Or/ 

m—o7i 

where  P  is  the  first  term  of  a  binomial,  Q  the  second  divided  by 

the  first,  —  the  index  of  the  power  or  root,  and  A,  B,  C,  &c., 

the  terms  immediately  preceding  those  in  which  they  are  first 
found,  including  their  signs  -j-  or  — . 

332.  This  theorem  may  be  applied  to  any  particular  case,  b}" 
substituting  the  numbers  or  letters  in  the  given  example  for  P 
Q,  m,  and  7i,  in  either  the  above  forrauh-c,  and  then  finding  the 
result  according  to  the  rule. 

When  the  index  of  the  binomial  is  a  whole  number,  the  series 
will  terminate,  as  observed  under  the  article  Involution  ;  but 
when  it  is  a  negative  or  fractional  number,  as  in  the  following 
examples,  the  series  will  proceed  on  ad  mjinitian,  and  will 
become  more  convergent  the  less  the  second  term  of  a  binomial  is 
with  resnect  to  the  first. 


>C    I    -  ^=x. 


•     -~->^   -    ■      J-    YX         '' 

312   ^  ALGEBRA. 


EXAMPLES.  .  iJ^. 

1.  It  is  required  to  convert  {ar-\-xY  into  an  infinite  series. 


X      071 

Let  P=aP,  Q=—,  —=h->  0^  7^i=l  and  n^2. 
a-    71 

m  TO 

Then  P^=(a2)~=(a2)2-^^^^^ 
71  '^     1     a-     2<z 

wz — 3?z  1 — 6        ^x"        X  3.5a;* 

?w— 4?z  1— 8  3.52;*        a;  _     3.5.72;^     _ 

~5^^    ^"""10"^  ~2X6:8Z^^^~'2.4.6.8.10a'-^~~ 

i. 
Therefore  (a"-fa;)^  = 

.X        a;2     .     32;^  3.52;*  _         3.5.72;^ 


'  2fl     2.4a-^  '  2.4.6a^     2.4.6.8a^  '  2.4.6. 8.10^'-^ 

The  pupil  will  readily  perceive  that  the  law  of  formation  of 
the  several  terms  of  the  series  is  sufficiently  evident. 

,,2 

2.  Required  the  development  of  — ^^ r-  in  a  series. 

{x'^-y)h 

Here  — ^ -^x'ix'—y)-^,  P=x\  Q=—K,  7n=:—l, 

[x'^—y)^  ^- 

and  ?z=2. 

mm  1 

Hence  P'~=(2;-)^=  {x')-h=-=A. 

7«  .  _  1  y        II        ^ 

71  x         x^     Ix^ 

^n  ^'^-      4      ^2^^^~^^-2:4^^~^* 


BINOMIAL     THEOREM.  313 

37^       ^  6      '^2.4x^'^     x~1AM~ 

^^^^^'^^^'  (^^3^=r+2;^+2:fc+2A^^+2^^^       ^^^• 

^'  ,   2/   ,    -r    ,    3.5y^    ,    3.5.7?/*    ,     , 

^-^'  (^^)l=^'+2i+2i?+2r4i?+2A(3i?+  ^" 

This  last  equation  is  obtained  from  the  former  by  multiplying 
each  term  of  the  equation  by  x^. 

3.  Required  the  cube  root  of  9. 

Here,  9'^=(8+l)* 

Therefore,  P=8,  Q=|,  m=l,  and  n=3. 

Whence,  P~=8"=8^=2=^. 

??i  .  ,.     1     ^     1        1        „ 
-^Q=3X2X^,=3.,.=B. 

772— 7i„„       1  —  3  11  1  ^ 

3?z       ^        9     '^     3.6.2*'^2'^~3.6.9.2'~ 

TTz— 371  1-9  5  1__       5.8       _ 

47^        ^~   12   ^3.6.9.2' ^•P~"~3.(5.9.12.2i'^~ 
Therefore, 

-     9^_94.J 1,5  5.8         , 

""^3.2^     3.6.2*~^3.6.9.2^     3.6.9.12.2^""^'      * 

4.  What  is  the  square  root  of  a-\-h  1 
Here,  — =h»  -P=«,  and  Q=-. 

»  71       2  c 

m  1 

Then,  P"=a-=^ 

7^  I      \      a      *la         2 


314  A  L  G  F.  1{  U  A  . 

m — n^,.      1 — 2     a~2b     h          a'^L"         a~2Jr     ^ 
--—BQ=~    X— -X-= o— = ft-=^- 

?n — 2n^^^     1 — 4         ar'iW     h      ^a~2p     a~^b"      ^ 
on  b  <S         a       4ba  lb 

,    ,m-Zii^     1-6     a-f^-'^     h     harW         ha-W     ^ 
And  -^_DQ=_^_x^-X-=3^  =  -  ^g-  =^. 

1.  J        ^-^3       ^t^2        ^1^3        5^-1-34 

Therefore,  (a  +  Z')-=a^  +  ^^ ^ — j— ^^ j^^,  &c. 

5.  What  is  the  cube  root  of  7  ? 

Ans.  2-g-^-3-g;-^4-3g  9  2^—3.(5.9.12.2^"'  '^^' 

2 

6.  Expand  (1 — aY  into  an  infinite  series. 

^     2a     2.3.^2     2.3.8.a'^        „ 
Ans.  l___-^^_^^^^^_,&c. 

7.  It  is  required  to  convert  j,  or  its  equal  {^-\-x)~\i 

(l4-2')5        . 

into  an  infinite  series.  -.    i 

X      ^x"      e.llr^       61L16^ 
^,,,.  i__4________-|-__^^_     ,  &c. 

-1 

8.  It  is  required  to  convert  {a—hY  into  an  infinite  series. 

1-/,      ^^       3^2        3.7.^'"  3.7.11Z'^  „     \ 

Am.  a  \^---^^-^^^^-^-^^^--,k..^ 

INDETERMINATE   COEFFICIENTS. 

833.  This  is  a  general  method  of  obtaining  a  series  from  frac- 
tions, and  other  expressions,  without  either  performing  the  division 
or  extracting  the  root. 

lluLE.  Assume  a  series  with  unknown  hut  constant  coeffi- 
cients of  X,  increasi7ig  or  decreasing  in  the  same  way  as  if 
the  operation  was  j)crformed  at  length  ;  then  make  this  scries 
equal  to  the  givc7i  expression,  and,  charing   the   equation  of 


INDETERMINATE     COEFFICIENTS.  315 

fractions^  bring  all  the  terms  to  one  side,  so  as  to  make  the 
equation  =  0  ;  next  make  the  first  term  of  the  coejjicients  of 
the  several  powers  of  x  each  =  0,  and  there  will  arise  as  many 
independent  equations  as  there  are  unknown  coefficients,  from 
which  their  values  may  he  found  and  substituted  for  them,  in  the 
assumed  series. 

EXAMPLES. 

1.  Let  it  be  required  to  expand  ;— ; —  into  a  series. 

b-j-x 

Assume  ——=A-{-Bx-}-Cx^-{-Dx"-\-&g.  ;    then,   multiplying 

both  sides  by  b-\-x,  and  transposing  a,  we  obtain  Ab — a-'r 
{Bb-{-A)x-\-{Cb-\-B)x^--{-{Db-{-C)x''-l-&c.  =  0,  an  equation  which 
must  be  true,  whatever  be  the  value  of  x.  Now,  making 
the  first  term,   and  the  eoeflncients   of  the  several   powers  of 

X,  each  =  0,  we  have  Ab — a=0,  or  A=j',    Bb-\-a=0,    or 

B=:f=~;    Cb+B^O,   or   C~=  +  ^^     Db+c=0,    or 

C         a 
D  =  -=— — ,  &c.     And,   substituting   these  values   of  A,  B, 
b  b 

C,    D,    &c.,    in   the   assumed   series,   we   get = --{- 

b-\-x      b     b" 

cx"    ax 

-jj — 7r+  ^^->    "^   which,    it   is   obvious,    that   the   signs    are 

alternately  -|-  and  — ,  and  the  exponents,  both  in  the  numer- 
ator and  denominator,  increase  continually  by  1,  that  of  x  in 
the  numerator  being  always  1  less  than  that  of  b  in  the  de- 
nominator. 

2 

2.  Expand  — — r into  a  series. 

a'-{-2ax — x- 

,      2x     5a;2     12:r3  ,     „ 

Am.    1 \ — ^4-,  &c. 

a       a-        a^ 

3.  Expand  /y/  {or — x-)  into  a  series. 


316  ALGEBRA. 

,    ^         ,     1+2:?;     . 

4.  Jiixpand  - — ' into  a  series. 

1 — X — X- 

Ans.  l4-32:+4:?;2+72;^+lla;4-|-18^-^+,  &c. 

This  is  a  recurring  series,  in  wtiicli  each,  of  the  coefficients, 
after  the  second,  is  the  sum  of  the  two  preceding  ones. 

5.  Expand  /s/  (1 — a)  into  a  series. 

'  2~'2:4~2A6~2X6:8"'2.4.6.8.10~'      ' 

1—x 

6.  Expand rr—  into  a  series. 

1 — za: — 6x" 

Ans.  l^x-\-bx''+12,x^-\-41x^-\-\'l\a^-\-mbx\  &c. 

7.  What  is  the  expansion  of  (a—b)^  ? 


^/       5        3^2        3.7^^  3.7.11^-^  \ 

^7Z5.  a  1^1  4^-4:8^2-4x12^3-4^^12.160'^"'  ^''- ; 

8.  It  is  required  to  expand  [a-\-x)~^. 

1      2x  ,  32;2    4x^ 
Ans.  -5 r-^ — 2 r+,  &c. 

9.  It  is  required  to  expand  ^^j^^'  icx.-hl^ 

1      62:  ,  24x^^     mx^    „ 

Ans.  -o— — ri -r T-,  &c. 

9 

'  10.  It  is  required  to  find  the  expansion  of 


{c^xf 

2     4x  ,  Qz^     Sx^  ,     , 
^^^-  72—73+-^—^+'  &c. 


11.  It  is  required  to  find  the  expansion  of — rTTTTs' 


,       1/'       6/^  ,  24^*2     8O33  ,     ^ 
Ans.  -   1 -j ^ — H,  &c. 


12.  What  is  the  value  of r  in  a  series  ? 

1      3;       3x^       3.52;3        3.5.7a;* 
•   "^     2^.3+2:4^5    2X6^^"^2.4.6.8^«"~'  ^'^' 


SUMMATION     OF     SERIES.  817 

SECTION   XXXVI. 

SUMMATION   AND    INTERPOLATION    OF   SERIES. 

Art.  S34.  The  Summation  of  Series  is  the  method  of  finding 
a  terminated  expression  equal  to  tl\e  whole  series. 

Interpolation  is  the  method  of  finding  any  term  of  an  infinite 
series,  without  producing  all  the  rest. 

DIFFERENTIAL   METHOD. 

335.  The  Difi"erential  Method  consists  in  finding,  from  the  suc- 
cessive diflferenccs  of  the  terms  of  a  series,  any  intermediate 
term,  or  the  sum  of  the  whole  series. 

Problem  I. 

336.  To  find  the  several  orders  of  differences. 

Let  a-{-b-\-c-\-d-]-e-{-,  &c.,  be  any  series;  subtract  each  term 
from  the  one  following  it,  and  the  diff"erences  — ci-{-b,  — b-{-c, 
— c-\-d,  — d-\-e,  &c.,  will  form  a  new  series,  called  the  Jirst 
order  of  differences.  Again,  subtract  each  term  of  this  new 
series  from  the  one  that  follows  it,  and  the  differences  a — 2Z»-|-c, 
b — 2c-\-d,  c — 2d-\-e,  o;c.,  will  form  another  series,  called  the 
second  order  of  differences.  Proceed  in  like  manner  for  the 
third,  fourth,  fifth,  &c.,  order  of  differences,  until  they  at  last 
become  equal  to  0,  or  are  carried  as  far  as  is  required. 

337.  When  the  several  terms  of  the  series  continually  in- 
crease, the  differences  will  all  be  positive ;  but,  when  they 
decrease,  the  differences  will  -be  alternately  negative  and  posi- 
tive. 

1.  Required  the  several  order  of  differences  of  the  series  1,  6, 
20,  50,  105,  196,  &c. 

1,  6,  20,  50,  105,  196,  &c.,  the  given  series. 
5,  14,  30,     55,     91,  &c.,  1st  differences. 
9,  16,     25,     36,  &c.,  2d 
7,       9,     11,  &c.,  3d 
2,       2,  etc.,  4th 
0,  &c.,  5th 
27^ 


318  ALGEBKA. 

2.  Required  the  several  order  of  dififerences  of  the  series  of  P, 
2\  32,  42,  52,  &c. 

1,  4,  9,  16,  25,  &c.,  the  given  series. 
3,  5,    7,    9,  &c.,  1st  differences. 
2,    2,    2,  &c.,  2d 
0,    0,  &c.,  3d 

3.  Required  the  several  order  of  differences  of  the  series  of 
cubes,  l^  2\  3^  4^  5^  Ans. 

4.  Find  the  order  of  differences  in  the  series  ^,  ^,  -^,  j'^, 
^^2-,  &c.  ^?w. 

Problem  II. 
338.  To  find  the  first  term  of  any  order  of  differences. 

Let  d',  d",  d"\  d"",  &c.,  represent  the  first  terms  of  the  1st, 
2d,  3d,  4th,  &c.,  order  of  differences ;  then  d'z=z — a-\-b,  d"=a 
—2h-\-c,  d"'=-a-\-Sb—Sc+d,  d""=a—^b^Qc—U+e,  &c.; 
from  which  it  is  obvious  that  the  coefficients  of  the  several  terms 
of  any  order  of  differences  are  respectively  the  same  as  those  of 
the  terms  of  an  expanded  binomial,  and  are  obtained  in  the  same 
manner ;  for  the  terms  that  are  subtracted  are  actually  added, 
but  with  contrary  signs.     Hence  we  infer  that  d",  or  the  first 

difference  of  the  nth.  order  of  differences,  is  -\-a^nb-{-7i .  — ^ 

71 — 1   n — 2  -        0  -  .       1 .  1    o         11 

c^=-n. — ^— .  — p~"^±»  ^^'j  *o  n-{-l  terms  ;  m  which  formula  the 

upper  signs  must  be  taken  when  n  is  an  even  number,  and  the 
under  when  n  is  an  odd  number. 

5.  Required  the  first  of  the  fifth  order  of  differences  of  the 
series  6,  9,  17,  35,  63,  99,  148,  &c. 

Let  a,  h,  c,  d,  e,/,  &c.  =  6,  9,  17,  35,  63,  99,  148,  &c.,  and 
71=5.     Then 

n{n  —  l)       n{n—l){7i—'l)       n{7i—l)[n—'l){n—Z) 
-a  +  nb  ^       c-\  —         d  ~—  e 

^  ^(,,_l)(,,_2)(,^_3)(7^-4)  ^•^.^•'^•^^ 


2.3.4.5  •"  '  2      '    2.3 


SUMMATION,     ETC.,     OF     SERIES.  319 

494— 491=+3.     Ans. 

6.  Required  the  first  of  the  sixth  order  of  differences  of  the 
series  3,  6,  11,  17,  24,  36,  50,  72,  &c.  Ans.  —14. 

Problem  III. 

339.  To  find  the  wth  term  of  the  series  a,  b,  c,  d,  e,f,  &c. 

As  we  have  found  in  the  last  problem  that  d'= — a-\-b,  there- 
fore bz=a-\-d\  and,  in  the  same  manner,  wc  find  c=a-{-2d'-\-d", 
d=a-\-U'+M'-{-d"',  e=a^W-{-U"-{-W"-^d"'\  &c. ; 
whence  the  nth.  term  is 

n—1        n—1   n—2        ?^— 1   7i—2   n—S 
=ai~^d'Jr-^- .  -^d  -f--^  .  -^  .  -^-d"  +,  &c. 

7.  Required  the  7th  term  of  the  series  3,  5,  8,  12,  17,  &c. 

3,  5,  8,  12,  17,  &c.,  the  given  series. 
2,  3,    4,    5,  1st  difference. 
1,     1,     1,  2d  difference. 
0,    0,  3d  difference. 
Here  d'=2,  d"=\,  d"'=0,  and  n=7. 

Therefore  a-{-'^d'-\-'^  .  !^^"=3+^ .  2-f 

I^.llll?.  1=3+12+15=30=  the  7th  term. 

8.  Required  the  9th  term  of  the  series  1,  5,  15,  35,  70,  &c. 

A71S.  495. 

9.  Required  the  10th  term  of  the  series  1,  3,  6,  10,  15, 
21,  &c.  Ans.  55. 

Problem  IV. 

310.  To  find  the  sum  of  7i  terms  of  the  series  a,  b,  c,  d,  e,  &c. 
If  we  add  the  values  of  a,  b,  c,  &c.,  as  found   in  the  last 
problem,  we  obtain  2a-\-d'=a-\-b,  3a+3c^'+(^":=a+3+c,  4a+ 


320  ALGEBllA. 

Qd'-j-4d"+d'"=a-j-b-}-c-\-d,  &c.     "^ATierefore  it  is  evident  that 
the  sum  of  n  terms  must  be 

na+n.^-d-^n.^-.-^d  +n.-^.-^.-^d    +, 
&c. 

341.  When  the  differences  become  at  last  =  0,  any  term,  or 
the  sum  of  any  numbers,  can  be  accurately  found ;  but,  when  the 
differences  do  not  vanish,  the  formulae  in  this  and  the  preceding 
problem  give  only  an  approximation,  which  will  come  nearer  the 
truth  as  the  differences  diminish. 

10.  Required  the  sum  of  8  terms  of  the  series  2,  5, 10, 17,  &c. 
Here  n=S,  a=2,  d'=o,  d"=2,  and  d"'==S). 

Hence,  na-]-7i .  -^j-d'-^n  .  —-— .  ^d"=^ .  2-|-8 .  ^ .  3+ 
8. 1. 1. 2=16+84+112=212=  the  sum  of  8  terms. 

11.  Required  the  sum  of  100  terms  of  the  series  1,  2,  3,  4, 

5,  &c. 

Here  1,  2,  8,  4,  5,  6,  &c.,  given  series. 
1,  1,  1,  1,  1,  &c.,  1st  difference. 
0,  0,  0,  0,  &c.,  2d  difference. 
Here  ?2=100,  a=l,  and  d=X. 

n—\,     .^.  .  ..^    /lOO— 1' 


iia 


+,i.!i_^=100-|  100.(^1^^)  1=5050.     Am. 


12.  Required  the  sum  of  12  terms  of  the  series,  1,  4,  10, 
20,  85.  Ans.  1365. 

13.  Required  the  sum  of  n  terms  of  the  series  1-,  2-,  3-,  4-, 
h\  6-,  7-,  &c. 

Here  1,  4,  0.  16,  25,  36,  49,  &c.,  given  series. 
3,  5,    7,    9,  11,  13,  &c.,  1st  difference. 

2,    2,    2,    2,     2,  &c.,  2d  difference. 
0,    0,    0,    0,  &c.,  3d  difference. 


SUMMATION     OV     SERIES.  321 

Leta=l,  rf'=3,  and^"=2. 

Then  na-^--^-^ d'-\ ^ — a  = ^ . 

14.  Required  the  sum  of  7i  times  of  the  series 

1\  2\  3^  4^  5^  6-^  &c. ;  1,  8,  27,  64,  125,  216,  &c. 
Here  1,  8,  27,  64,  125,  216,  &c.,  given  series. 
7,  19,  37,    61,    91,  &c.,  1st  difference. 
12,  18,    24,    30,  &c.,  2d  difiference. 
6,      6,      6,  &c.,  3d  difiference. 
0,      0,  &c.,  4th  difiference. 

Let  a=l,  d'=l,  d"=l'2,  d'"=G.     Then 

^.  .  ^(^^-1)^.  ,  n{n-l){?i-2)^^,,  ,  ni7i-l){?i-2){n-S)^  ^^, 
"^        2         "^       2      :      3~    "^       2.3.4 

7n{n—l)     12?i{7i—l){7i—2)     Q7i(n—l){7i—2){n—3) 
=n-f- -; 1 ^ 5 1         n — 


o 


4 


=71 -j ^ \-2n^ — b?i- 4-471 -f 


2         '  '         '  4 

471     Un^-'Un     Sti"^— 24?i-+16?^     yz^— Gti^+Utz-— 67t_ 
T"'  4  *  4  '  4  "~ 

7i^+2;z"-f?z-     7i-(7i-fl)2  . 

, — ■ — = — ^^ — ■ ==  sum  01  n  terms,  as  required. 

4  4  .,  ^ 

15.  What  is  the  number  of  cannon-shot  in  a  square  pile,  the 
bottom  row  consisting  of  25  shot  "^  ?  -  Ans.  5525. 

16.  I  have  10  square  house-lots,  whose  sides  measure  5,  6,  7, 
8,  9,  &c.,  rods,  respectively.  What  is  their  value,  at  25  cents 
per  square  foot  ?  A?is.  $67,041,561. 

*  Shots  and  shells  are  generally  piled  in  three  diflferent  forms,  called 
triangular,  square,  or  oblong  piles,  according  as  their  base  is  either  a 
triangle,  a  square,  or  a  rectangle. 

A  square  pile  is  formed  by  the  continual  laj'ing  of  square,  horizontal 
courses  of  shot,  one  aboye  another,  in  such  a  manner  as  that  the  sides  of 
their  courses  decrease  by  unity  from  the  bottom  to  the  top  row,  which 
ends  also  in  one  shot. 


/ 

y 

1 

€ 

'/ 

■\ 

ALGEBRA 

. 

822  '     '' 

17.  There  are  5  cubical  blocks  of  marble,  whose  sides  meas- 
ure, respectively,  2,  3,  4,  5,  and  6  feet?  What  is  their  value 
at  $2.75  per  cubic  foot  ?  Am.  $1210. 

18.  What  is  the  number  of  shot  in  a  square  pyramidical  pile, 
whose  side  at  the  base  contains  100  shot  ?  Ans.  338350. 

19.  What  is  the  sum  of  20  terms  of  the  series  1^,  2'^  3",  4^ 
5^  6^  &c.  ?  Am.  44100. 

20.  What  is  the  sum  of  20  terms  of  the  series  l^  2*,  3^  4^ 
h\  OS  &c.  ?  Am.  722666. 

Problem  Y. 

312,  To  find  a  fraction  that  will  express  the  value  of  a 
geometrical  series  to  infinity. 

In  Art.  284  we  find  that  the  sum  of  an  infinite  series  is 
obtained  by  the  following  formula : 

c        ^ 
1 — r 

and,  by  this  formula,  we  may  find  the  sum  of  algebraic  series. 

EXAMPLES. 

1.  What  is  the  sum  of  the    series    \-\-a-\-a-'-\-d'-\-a''^  &c., 

carried  to  infinity  ?  ,  1 

^  Am 


\—a 

By  the  above  formula,  the  first  term  of  the  series  will  be  the 
numerator  of  the  fraction,  and  the  denominator  is  obtained  by 
subtracting  the  second  term  from  the  first. 

2.  What  fraction  will  express  the  exact  value  of  the  series 

14-5+25  +  125,  &c.,  to  infinity  ?  1 

Alls. 


1-5* 

3.  What  fraction  will  express  the  infinite  series  1 — a-\-(r — a'' 
-\-a^—a',  &c.  ?  ,1 


l+<2 
A      -itn  n  •  .        ll       l)h        Irh 

4.   What  fraction  will  express  the  series  — ! — zr-\ — ^+>  &<?•>  to 

^  a     a^       a 

<                         Ji 
infinity?  :;  Ans.    -. 

/h  ./       --,  . 

•< —  -  11.    <(L^.^ 


SUMMATION     OF     S  E  R  I  F,  ;>  T  323 

.         1111 

5.  What  is  the  sum  of  the  series  — j — --j — ^-j — -,  +  ,  etc.,  to 

X     X"     x"     x^ 

infinity  ?  Atis 


x—\ 

6.  What  fraction  will  express  the  series  l  +  24-4-[-5^+16,  &c., 

to  infinity  ?  Ans.     — — . 

1     x'^     x:^     x!' 

7.  What  fraction  is  equal  to  the  series ^-| — ; =+,  &c., 

a     a'     a'     a' 

.  ,  ,  Gj 

to  infinity  ?  Ans 


ar-\-x'' 

8.  What  fraction  wiU  express  the  value  of  l-j-l  +  l+l?  &c., 

-I 

to  infinity  ?  Ans.    z. — r-. 

O  o 

X"      X 

9.  Express  by  a  fraction  the  value  of  the  series  x4-'—-\-  — 

-f--:,4-)  &c.,  to  infinity.  Ans.   — '-■-. 

a'  a — X 

X       X"       X," 

10.  What  is   the  value   of  the  series  1 \-—,——-\-^  &c., 

a     a~     a' 

to  infinity  ?  Ans.    . 

''  a~{-x 

111 

11.  Required   the  sum  of  the  series   t^4"4To4"^7t4"'  <^^-' 

i.A     l.o     0.4 

continued  to  infinit3^  Ans.  1. 

This  question  may  be  performed  by  separating  the  factors  of  the 
denominators  so  as  to  form  two  series,  and  then  subtractinrr  the 
less  from  the  greater,  as  follows : 

Let    l-|-2~j~3~{~4'  ^^'  =  '^^  greater  series. 
And  2~f"3~l~4»  ^^-  =  ^^®  -^^^^  series. 

Then  1  =      the  sum  of  the  series. 
XoTK.  —  Another  method  may  be  found  in  the  Key. 

12.  Required  the  sum  of  the  series  =— r-f— r-^4--^^+T-^4-j 

^  l.i~2.o'  3.0^4.7  ' 

&c.,  to  infinity.  Ans.  -^. 


324  ALGEBRA. 

SECTION   XXXVII. 

CUBIC   EQUATIONS,    CONTAINING    ONLY   THE   THIRD    AND 
SECOND    POWERS. 

Art.  343»  Any  numerical  equation,  containing  only  the  third 
and  second  powers  of  the  unknown  quantity,  and  having  one 
rational  root,  may  be  reduced  by  rendering  both  of  its  members 
perfect  squares,  and  extracting  the  square  root  of  both  sides : 
completing  the  operation  by  former  rules.  The  only  difficulty 
lies  in  multiplying  the  equation  by  such  a  number  that,  after 
adding  to  each  side  the  fourth  power  of  the  unknown  quan- 
tity, and  the  second  power  with  a  coefficient  easily  determined, 
both  sides  will  be  perfect  squares.  This  multiplier  must  be 
ascertained  by  trial ;  for,  though  a  general  formula  might  be 
given  for  obtaining  it,  yet  it  would  be  so  complicated  as  to  be  of 
no  practical  use.  It  may  be  either  an  integer  or  a  fraction,  and 
is  positive  or  negative  according  to  the  sign  of  the  known 
quantity. 

Though  there  always  is  such  a  multiplier  whenever  the  un- 
known quantity  has  one  rational  value,  yet,  when  the  numbers 
are  very  large,  or  the  equation  is  very  complicated,  it  may 
not  be  readily  found,  and  the  process  of  trial  may  become  too 
tedious  to  be  of  service.  Whenever  the  equation  does  not  con- 
tain too  large  numbers,  the  pupil  will  find  little  difficulty,  if  he 
thoroughly  understands  the  following 

Rule.  Divide  both  sides  of  the  equation  by  the  coefficient  of 
the  unknown  cube,  if  it  have  any  expressed.  Place  the  third 
power  of  the  unknown  quantity  on  one  side  of  the  equation,  and 
the  second  power,  with  the  known  quantity,  on  the  other.  Mul- 
tiply both  sides  by  the  number  nearest  to  unity  which  will  7)iake 
the  known  quantity  a  positive  square;  or,  which  is  the  same 
thing,  separate  the  known  number  into  two  factors,  one  of  which 
shall  be  the  greatest  square  contained  in  it.  and  multiply  both 
sides  by  the  other  factor. 

Multiply  the  last  equation  by  4  ;  add  the  fourth  power  of  the 


CUBIC     EQUATIONS.  325 

unknown  quantity^  and  the  second  power,  with  a  coefficient  equal 
to  the  square  of  half  the  coefficient  of  the  third  power,  to  each 
side  ;  and  extract  the  square  root  of  both  sides,  if  possible.  By 
taking  like  signs  of  the  two  members  of  the  equation  in  evolving, 
we  shall  obtain  one  root ;  and,  by  taking  unlike  signs,  the  other 
two  may  be  found  by  quadratic  equations. 

But,  if  that  member  of  the  equation  which  contains  the  knoion 
quantity  is  not  a  perfect  squure,  substitute  1,  9,  16,  f ,  ^,  \,  ^, 
or  some  other  square  number,  in  the  place  of  4,  a7id  proceed 
as  above,  till,  by  trial,  a  number  is  found  which  will  accomplish 
the  object. 

Note. —  1.  The  sum  of  the  three  values  of  the  unknown  quantity  should 
always  be  equal  to  the  coefficient  of  the  second  power  in  the  original 
equation,  after  dividing  by  the  coefficient  of  the  cube,  and  placing  it  on 
the  same  side  with  the  known  quantity,  opposite  the  positive  cube  ;  hence, 
if  two  values  were  known,  the  other  might  easily  be  found. 

2.  When  one  of  the  values  is  known,  the  others  might  be  found  by  the 
usual  method  ;  bringing  all  the  terms  of  the  original  equation  to  the  same 
side,  and  dividing  by  the  difference  between  the  unknown  quantity  and 
its  known  value,  reducing,  by  quadratic  equations,  the  equation  thus  pro- 
duced. But  the  three  values  are  here  given  directly,  by  using  the  different 
signs  in  evolving,  thus  rendering  the  solution  shorter,  and  more  satisfac- 
tory. It  is  evident  that,  in  extracting  the  square  root  of  an  equation, 
both  sides  may  be  considered  positive,  or  both  negative,  or  either  one 
positive  and  the  other  negative.  Thus  the  square  root  of  the  equation 
4a2_8a6+462=c2-f-2cc?-f(^^  is  -f(2a— 26 )=-[-( c-{-^),  or  —{2a— lb) 
=—{c-\-d),  or  -i-(2a— 26)=— (c+c^),  or  _(2a— 26)=-f  (c-f-^i).  But, 
if  both  sides  take  like  signs,  the  result  will  be  the  same,  whether  they  are 
both  positive  or  both  negative,  as  the  signs  of  both  sides  of  an  equation 
may  always  be  changed;  while,  if  they  take  unlike  signs,  a  different  equa- 
tion will  be  produced,  it  making  no  difference  which  side  is  positive. 
Hence,  there  are  but  two  results  that  can  be  obtained,  and  we  have  pre- 
ferred to  express  them,  in  the  following  examples,  by  the  same  method  as 
in  quadratic  equations  ;  prefixing  the  sign  i;  to  the  right-hand  member 
of  the  equation  produced  by  evolution. 

3.  By  observing  whether  the  root  of  the  known  quantity  is  greater  or 
less  than  half  the  coefficient  of  the  second  power  on  the  same  side,  if  we 
also  notice  the  sign,  we  may  usually  know  whether  the  multiplier  we  have 
used  is  too  small  or  too  large.  When  there  are  two  rational  values  of  the 
unknown  quantity,  of  course  the  third  will  be  rational,  and  there  will  be 
three  different  multipliers,  which  will  answer  our  purpose,  thus  giving 
three  different  solutions  for  the  same  example. 

28 


826  ALGEBRA. 


EXAMPLES. 


1.  What  are  the  values  of  x  in  the  equation  7? — a:^=4  ? 

Here  the  multiplier,  which  would  make  the  known  quantity  a 
perfect  square,  is  unity ;  therefore  we  transpose,  and  multiply 
by  4,  42;3=42;2-j-16. 

/4\2 
Adding  x'^  and  (  -  J  a:^  to  each  side,  a;^-f"4^+4a;^=^*"l~^^^"t"l^' 

Evolving,  a;2+2a^=:±(a:^+4.) 

Taking  the  positive  sign  and  cancelling,  22:= 4. 

Dividing,  a: =2. 

Taking  the  negative  sign,  7?-\-^x:=. — :ii? — 4. 

Transposing  and  dividing,  a:^-{-a;= — 2. 

By  quadratics,  xz=. . 

Hence  a:=2,  or .     Ans. 


The  sum  of  their  values,  2 -J j — . — ,  is  1. 

2.  What  are  the  values  of  a;  in  the  equation  4a;^-|-10a:^=9  ? 

Conditions,  4a:^-j-10a;^=9. 

5        9 

Dividing  by  4  and  transposing,  a:^= — h^^+t* 

9 

-  being  a  square,  multiply  by  4,  4a:^= — 10a;^-{-9. 

Adding  x"  and  i-\  x\  x'^^A7?-\-^x''=x''—Qxr-\-^, 

Evolving,  a;2+2a:=±(a:2— 3). 

Taking  the  positive  sign  and  cancelling,  22:= — 3. 

3 

Dividing,  x:^ — -. 

Taking  the  negative  sign,  x'^-\-2x=sz—x'^-\-Z. 


CUBIC     EQUATIONS.  327 

3 

Transposing  and  dividing,  x--\-x=-, 

A 

Bj  quadratics,  xz=. w^- 

A 

Hence  2;=—-,  or .     Am. 

The  sum  of  these  roots  is . 

2 

3.  Given  32;^— 2:^2^931  to  find  the  values  of  a:. 
(1.)  Conditions,  3r''— 2:^2=931. 

2^2  .  931 


3  "^  3  * 


(2.)  Dividing  and  transposing,  0? 

/Q  X  rru  .  .     931  .    49 

(d.)  ihe  greatest  square  m  -5-  is  — - ; 

o         1 

^,       „  931     49     19 

therefore  ___=__X— . 

Multiplying  (2)  by  -^, 


Multiplying  by  4, 


3'  3   ~~    3     '       9     • 

76z'^_152a;2     70756    ' 
3  ""~3      '      9~* 


Adding  a;''  and  ^--j  x%  2;^+ _^-{— ^—=2:^4— ^4-_^_. 
17    1  •  ,  ,  38a:         /  ,     266\ 

Evolving,  x''-\-—=:±(x'  +  —). 

Taking  the  positive  sign  and  cancelling,  38:c=266. 

Dividing,  x=7. 

m  T      .u  r        •  o  ,  38^  ,     266 

Taking  the  negative  sign,  ci^-\ — —z=—x- ^. 

3  3 

Transposing  and  dividing,  ar-| — j— = — -^. 

o  o 

■R          ;i    .•                                                        — 19±V  — 1^35 
J3y  quadratics,  x= . 

6 

XQ.^,^ 1*^35  2 

Hence  x=7,  or =^ — .     The  sum  of  these  is  tt. 

o  3 


328  ALGEBRA. 

4.  Given  a;^=12a;^— 81  to  find  the  values  of  ^. 
Conditions,  a;^=12:r2_8i. 

By  multiplying  both  sides  by  —1,  — 81 

becomes  a  positive  square,  — a;^= — 122;^-|-^1« 

We  find  that  neither  4,  9,  16,  nor  25, 

will  answer  our   purpose,   and  we 

multiply  by  36,  —  36r^=— 432a;24-2916. 

Adding  a;4  and  (--\  x\    z''—m3?-\-Z2^^=zx^—\^^x'-{-2^1Q. 

Evolving,  a;2_i8^__j_(2;2_54). 

Taking  the  positive  sign  and  cancelling,  — 18^:= — 54. 
Changing  signs  and  dividing,  a: =3. 

Taking  the  negative  sign,  x^ — 18a;= — x^-\-b^. 

Transposing  and  dividing,  a? — 92;=27. 

Completing  the  square,  x^^9x-]—^z=27-\—j-=—^. 

Evolvmg,  ^""2^~~2~* 

Transposing,  .=^^. 

9rh3A/2l 
Hence  a;=3,  or .     The  sum  of  these  is  12. 

5.  Given  x'-^-x^^z — 4  to  find  the  value  of  :r. 

Ans.  —2,  or  — . 

•  6.  Given  7x^=x'^-\-^Q  to  find  the  values  of  x. 

Ans.  x=Q,  or  3,  or  — 2. 

7.  Given  x^—4:X^= — 9  to  find  the  values  of  x. 

Ans.  x=S,  or ^ . 

2 

8.  Given  2:c^=99— 5:^2  to  find  the  values  of  a:. 

Ans.  x=6,  or . 

4 

9.  Given  4:c"+102;-=125  to  find  the  values  of.?:. 

A?is.  x=2i,  or  i:!±|v^. 


CUBIC     EQUATIONS.  329 

10.  Given  a:^=8a;^-}-363  to  find  the  values  of  x. 

Ans.  a:=ll,  or . 

11.  Given  372;-=7a;^-|-144  to  find  the  values  of  a:. 

Ans.  x=4:,  or  3,  or  — If. 

CUBIC   EQUATIONS   CONTAINING   ONLY   THE   THIRD   AND   FIRST 

POWERS. 

Art.  344,  Any  numerical  equation  containing  only  the  third 
and  first  powers  of  the  unknown  quantity,  and  having  one 
rational  root,  may  be  reduced  by  multiplying  both  sides  of  the 
equation  by  the  unknown  quantity,  and  adding  the  second  power 
to  each  side,  with  such  a  coefiicient  as,  after  adding  a  number 
readily  determined,  will  make  them  perfect  squares.  The  only 
difficulty  lies  in  finding  this  coefficient,  which  must  be  ascer- 
tained by  trial ;  though,  by  adopting  the  following  rule,  it  can 
readily  be  found,  unless  the  equation  is  so  complicated,  or  the 
numbers  so  large,  as  to  render  the  operation  tedious. 

In  this  and  also  the  preceding  case,  the  rule  might  perhaps  be 
so  framed  as  to  obtain  the  roots  without  reducing  the  coefficient 
of  the  cube  to  unity,  the  two  methods  bearing  somewhat  the 
same  relation  to  each  other  as  the  two  in  quadratic  equations. 
But  we  have  preferred  to  use  fractions  occasionally,  rather  than 
render  the  rule  more  complicated. 

Rule.  Divide  both  sides  of  the  equation  by  the  coefficient  of 
the  unknown  cube^  if  there  be  any  expressed.  Place  the  two 
powers  of  the  loihnmvn  quantity  on  one  side,  and  the  hncnvn 
quantity  on  the  other,  and  multiply  both  sides  by  the  unknown 
quantitij  ivith  such  a  sign  as  shall  render  the  fourth  power 
positive. 

Separate  the  coefficient  of  the  first  power  of  the  unknown 
quantity  in  the  equation,  thus  produced,  into  two  factors,  and  add 
the  second  power,  with  a  coefficient  equal  to  tJie  square  of  one  of 
these  factors,  usually  the  smaller,  to  each  side.     If  it  make  the 

*  Until  the  factors  are  found,  it  is  sometimes  better  to  give  the  known 
quantity  and  the  first  power  a  common  denominator,  even  though  the 
former  might  be  reduced  to  a  whole  number. 

28=^ 


330  ALGEBRA. 

coefficient  of  the  square,  on  the  same  side  as  the  fourth  power, 
equal  to  the  other  factor,  add  the  square  of  half  this  coefficient  to 
each  side,  and  extract  the  square  root  of  hoth  members,  completing 
the  operation  by  former  rules. 

But,  if  the  above  coefficient  be  not  equal  to  the  other  factor, 
separate  the  same  number  into  two  other  factors,  or  perhaps  ex- 
change  the  same,  and  proceed  in   the  same  way  till  the  right 

ones  are  found. 

« 

Note  1.  — The  sum  of  the  three  values  of  the  unknown  quantity  should 
always  be  0,  as  there  is  no  second  power  in  the  original  equation  ;  hence, 
if  two  are  known,  the  third  will  be  equal  to  their  sum,  with  the  sign 
changed  ;  and  there  must  always  be  one  positive  and  one  negative  value, 
the  other  being  sometimes  positive  and  sometimes  negative. 

2.  We  obtain  the  three  values  by  the  same  method  as  in  the  preceding 
case,  prefixing  the  sign  -j-  to  the  right-hand  member  of  the  equation  in 
evolving.  Taking  the  positive  sign,  we  obtain  either  one  or  two  values,  and 
the  negative  sign  gives  the  remaining  values  or  value.  When  one  of  the 
values  is  known,  the  others  might  also  be  found  by  bringing  all  the  terms 
of  the  original  equation  to  the  same  side,  and  dividing  by  the  difference 
between  the  unknown  quantity  and  its  known  value. 

3.  When  two  of  the  values  are  rational,  the  third  will  of  course  be 
rational ;  and  there  may  be  three  different  methods  of  separating  into 
factors,  each  of  which  will  answer  the  purpose,  thus  giving  three  different 
solutions  of  the  same  equation. 

EXAMPLES. 

1.     Griven  o;^ — %x=l  to  find  tlie  values  of  a;. 

Conditions,  o? — 3a;=2. 

Multiplying  by  x,  x'^ — Sx"=2x. 

Separating  the  coefficient  of  2:?;  into  factors,  2x1' 
Adding  (iyx"^  to  each  side,  '  x^—2x^=x'^-\-2x. 

Add  (1)2,  x'—2x''-{-l=x''-\-2x-^l 

Evolving,  x'^—l=±:{x+l). 

Taking  the  positive  sign,  and  transposing,  x^ — :r =2. 
By  quadratics,  x=2,  or  — 1.    A?is. 

The  sum  of  these  is  1 ;  hence  the  other  value  is  — 1,  and  the 
equation  has  two  equal  roots,  — 1,  and  — 1. 


CUBIC     EQUATIONS.  331 

2.  Given  l()x=x^-}-S  to  find  the  values  of  x. 
Conditions,  10x=3:^-\-d. 
Transposing,                                      — ci^-{-10x=B. 
Multiplying  by  — x,                          x* — 10x^=—Sx. 
Separating  into  factors,  3=3 Xl* 
Adding  (3)'^;^  to  each  side,                     x^ — x'=dx'^ — dx. 

Since  coef.  of  x^  =  the  other  factor, 

add  {±y%  x'—x^-i-i=9x''—dx+l. 

Evolving,  x^-—}=±{dx—^). 

Taking  the  positive  sign,  and  cancelling,     x-=Sx. 

Dividing,  x=S. 

Taking  the  negative  sign,  x^ — ^= — 3z-j-J. 

Transposing,  x'^-\-dx=l. 

By  quadratics,  .==^^ . 

Hence,  a;=3,  or .     The  sum  of  these  is  0. 

3.  Given  4r'-|"^^=l^^  *o  find  the  values  of  x. 
Conditions,  4^:3+3^=182. 

Dividing  by  coefficient  of  7?^  ^^~1""T^^~T~* 

Multiplying  by  x,  a;*-[-_=__. 

.      182.       ^   ^  182     14     ^„ 

Separating  —^  into  factors,  ~r^^~T^^^' 

Adding  {-t-)^"  to  each  side,  a;^-f-132;^=:— j — j — -— . 

Since  coefficient  of  x^  =  the  other  factor, 

add  [-^^  ,  ^•^+132;-+(^-j  =_+_+^_  ) 


i3y 


Evolving,  ar+— =±(^-2-H-j. 

7.r 
Taking  the  positive  sign  and  cancelling,      2;-=-^. 


832  ALGEBRA. 

Dividing,                                                         ^=o- 
Taking  the  negative  sign,  a;^-|— — -= — — •. 

Transposing,  rc2_|_     __ — \'^ 

7x    49  49         159 

Completing  the  square,  a;2_|_      i      ___][3_|_     _. — 

^    ,  .                                                         ,7         V"^=T5^ 
Evolving,  ^+|=zfc ^ . 

^             .                                    *                       _-7±aA15^ 
Transposing,  :c=: r . 

The  sum  of  these  values  is  0. 

4.  Given  x^—7x=:Q  to  find  the  values  of  a:. 

Atis,  x=S,  or  — 1,  or  — 2. 

5.  Given  :r^=37a:-|-84  to  find  the  values  of  x. 

Ans.  x=7,  or  — 3,  or  — 4. 

6.  Given  2x^-\-7x=z4:74:  to  find  the  values  of  x. 

Ans.  x=Q,  or  ^ . 

7.  Given  9a;3=169:i:+280  to  find  the  values  of  a:. 

7  8 

Ans.  xv=o,  or  — ^r,  or  — -. 

o  o 

8.  Given  x^ — 3:r=322  to  find  the  values  of  a:. 

Atis.  x=z7,  or . 

Problems. 

1.  There  is  a  cubical  block  of  marble ;  and  if  50  be  added  to 
the  number  of  square  feet  in  half  its  surface,  it  will  be  equal 
to  the  number  of  cubic  feet  in  its  contents.  What  are  the  solid 
contents  of  the  block  ? 

Let  X  =  the  side  of  the  cube. 

Then,  x^  =  the  contents  of  the  block. 

And  x^  =  the  superficial  contents  of  one  side  of  the 

block. 


CUBIC     EQUATIONS.  333 

Then,  Zx^  =  the  superficial  contents  of  one-half  the 

surface  of  the  block. 

Therefore,    :r^=3a;2-|-50. 

Multiplying  both  sides  by  8,                  82r^=24:c2_j_4oo. 

Adding  x"^  and  square  of  4:r,  x'^-{-  ^x^-{-lQx^—x^-\-A.^x^-\-A.^^. 

Evolving,  a;2-f4;^:=2;2^-20. 

Cancelling,  .4x=20. 

Dividing,  a:=5. 

Therefore  the  contents,  a;^=125  cubic. 

2.  A  gentleman  having  asked  a  lady  her  age,  she  replied, 
that  if  29  times  the  square  of  her  age  were  subtracted  from 
twice  its  cube,  the  remainder  would  be  225.  What  was  the 
lady's  age  ? 

Let  a:=  lady's  age. 

Then,  2r'— 29a;2=225. 

Transposing,  23?=2^x'^J^  225. 

Adding  :c'*  and  the  square  of  ^,  x'^-\-2x^-{-x^z=x'^-\-^{)x^-\-21b. 
Evolving,  x^-\-x  =  a;^  -f  1 5 . 

Cancelling,  x=lb  years. 

3.  A  boy,  being  asked  what  he  gave  for  his  books,  replied, 
that  if  51  times  the  square  of  the  number  of  dollars  he  gave*  for 
them  were  subtracted  from  6  times  the  cube  of  the  number,  the 
remainder  would  be  900.     What  was  the  price  of  the  books  ? 

A71S.  $10. 

4.  A  man,  being  asked  how  many  dollars  he  had  in  his  pockets, 
replied,  that  if  three  times  the  cube  of  the  number  he  had  in 
his  pockets  were  added  to  five  times  the  square  of  the  number 
which  he  had,  he  should  have  272.  Kequired  the  number 
he  had  in  his  pockets.  Ans.  $4. 

5.  A  boat  has  been  sailing  two  hours,  with  a  light  breeze, 
against  a  strong  current ;  nineteen  times  the  number  of  miles  it 
has  sailed  is  equal  to  the  cube  of  that  distance,  added  to  thirty 
miles.  How  far  has  it  sailed  ?  Ans.  It  has  gained  either  3 
miles  or  2  miles,  or  it  has  lost  5  miles. 


334  ALGEBRA 


MISCELLANEOUS   QUESTIONS. 

1.  Multiply  7a/^T3V^— V^"'  by  Oa//.  Ans. 

2.  Multiply  a^-i-P  by  a-^—b-\  Ans.  ar'^W—cC-h-''. 

3.  Multiply  ar-\-}f  by  «-2«+^-". 

4.  Multiply  tsfax  by  — a/«^.  Ans.  — ax. 

5.  Divide  — a  by  — 3a.  J.7i5.  \. 

6.  Divide  a"""*  by  oJ",  Am.  a-"". 

7.  Divide  a^-\-z^  by  a+^'     ^^^'  ^"^ — a^x-}-a^x^ — az^-\-x*. 

8.  Multiply  y'^-j-x'"  by  2/—^. 

-4?Z5.  2/"'+^+a;'"2/— ^2/'"— ^'""^^. 

_    -^  .  .  ,     1       wfl   .    wV  ,  -  ,  wV:c"~^ 

9.  Multiply  -^—^-^^^^  by  :r''+7ia2:"-^4—^— . 

Ans.  l+-r^. 

10.  Divide  1 — x^  by  1—x. 

Ans.  lJ^x-^x^+x^-\-x^+a^-\-x^-\-x^, 

11.  Multiply  3/^G>=a^  by  4VF=^. 

^7z^.  12^'(^— 3a:c«— 2aV+3aV+6aV+a^a:^— a^a;2— 6aV 
— 3aV+2a^2:+3a'':r2— «^) 

„    _.        41-35Z       7-2:^2      l+3a:     2:^-2^  ,    .   ,  ,, 

12.  Given  — -7- — — =-_i_ __ ^  to  find  the 

105         14(0;— 1)        21  6 

value  of  a:.  Ans.  a:=:4. 

13.  Given  a/x-\-9=1-{-/s/x  to  find  the  value  of  a:. 

Atis.  z=16. 

14.  Given  .^ — — r^=l — = — itt^ — r-r^^  to  find  the  value 

1— 2z     7— 2a:  7— 16z+4a;- 

of  a:.  Ans.     2;=— |^. 

15.  Given  (A/^+28)(A/FF^)=(V^+3t3)(A/^+¥)  to  find 
the  value  of  x.  Atis.  a;=4. 


y^\  ~  ^ 

~   '}l.  ^    '       MISCELLANEOUS     QUESTIONS.  335 

16.  Qiven  {x — l)/s/''lx—x^=-^  to  find  x. 

Ans.  x= 7-^—. 

17.  Given  x—2/s/ x+2=\-\-,^ x^—6x-{-2  to  find  x. 

Ans.  2:=9±4VT,  or  ^ — . 

18.  Given  aJ a-\-x=l^ x^ — bax-{-b^  to  find  the  value  of  a;. 

Ans.  x=—- — . 
7a 

19.  Given  P=a^-\-bx  to  find  the  value  of  x. 

Ans.  x= — - — . 

0 

20.  Given  ^{x—a)—l{2x—U)=l^a-\-llb  to  find  the  value 
of  a:.  Am.  a;=25a-{-243. 

21.  Given  i^+-^,+^?^±l?=5£f+^   to  find   the 

value  of  a:.  .  ab 

Am.  X 


a-\-b 

22.  Given  (^+3:F+(^-^)'  ^^^  ^^  ^^^^  ^j^^  ^^1^^  ^^^^ 

(<2-|-^)  — (a— a:)2" 

2a3^ 

Ans.  x=- — -. 

1+3 

23.  Given  ^  ^^  ^   _|_^^    -^  L^^^t  to  find  the  value  of  a:. 

x'^  x^ 

Am.  2;=4(a — 1). 

24.  Given  ^-+3^— (-— 3V=c-  to  find  the  value  of  2:. 

Ans.  x=- 


25.  A  gentleman  travelled  252  miles.  The  first  day  he  rode 
4  miles,  the  last  128,  and  each  day's  journey  was  double  the 
preceding  one.     How  many  days  was  he  performing  the  journey  ? 

Atis.  6  days. 

26.  A  gentleman  dying  left  his  sons  an  estate  of  $13,187.50. 
He  bequeathed  to  his  youngest  son  $1000,  to  the  oldest  $5062.50, 


836  A  L  G  E  B  11  A . 

and  ordered   that  each  son's  portion  should  exceed  the  next 
younger  by  the  ratio  of  1^.     How  many  sons  had  he  ? 

A71S.  5  sons. 

27.  The  first  term  in  a  geometrical  progression  is  3,  the  last 
term  ^,  and  the  sum  of  the  series  4f .  What  is  the  number  of 
terms  ?  Ans.  4. 

28.  The  first  term  in  a  geometrical  series  is  ^,  the  ratio  7,  and 
the  last  term  3361f .     What  is  the  number  of  terms  ?   Ans.  6. 

29.  What  are  the  three  arithmetical  means  between  -^  and  ^  ? 

30.  Required  the  sum  of  200  terms  of  the  series  1,  3,  5,  7, 
9,  &c.  Am.  40,000. 

31.  The  first  term  of  an  arithmetical  series  is  — 7,  the  tenth 
term  is  12.     What  is  the  sum  of  the  series  ?  Atis.  25. 

32.  If  a  man  travel  20  miles  the  first  day,  and  15  miles  the 
second,  and  so  continue  to  travel  5  miles  less  each  day,  how  far 
will  he  have  advanced  on  his  journey  the  8th  day  ? 

Ans.  20  miles. 

33.  The  first  term  of  an  arithmetical  series  is  5,  the  number 
of  terms  20  ;  what  must  the  coromon  difi'erence  be,  that  the  sum 
of  the  series  shall  be  123^-  ?  Ans.  -^Yu- 

34.  If  a  man  travel  20  miles  the  first  day,  19  the  second  day, 
I82V  tbe  third  day,  and  so  on  in  a  geometrical  progression,  in 
how  many  days  will  he  have  travelled  400  miles  ?         Ans.  ^. 

35.  A  merchant,  having  mixed  a  certain  number  of  gallons 
of  wine  and  water,  found  that  if  he  had  mixed  6  gallons  more  of 
each,  there  would  have  been  7  gallons  of  wine  to  every  6  gallons 
of  water ;  but,  if  he  had  mixed  6  gallons  less  of  each,  there 
would  have  been  6  gallons  of  wine  to  every  5  gallons  of  water. 
How  much  of  each  did  he  mix  ? 

Ans.  78  gallons  of  wine  with  66  of  water. 

36.  A  person  bought  2  cubical  stacks  of  hay  for  £41 ;  each 
of  them  cost  as  many  shillings  per  solid  yard  as  there  were 
linear  yards  in  a  side  of  the  other,  and  the   greater  occupied 


MISCELLANPOUS     QUESTIONS.  837 

9  square  yards  of  ground  more  than  the  less.  What  was  the 
price  of  each  ?  Atis.  £25  and  £16. 

37.  A  certain  man  owes  $1000.  What  sum.  shall  he  pay 
daily,  so  as  to  cancel  the  debt,  principal  and  interest,  at  the  end 
of  the  year,  reckoning  simple  interest  at  6  per  cent.  ? 

Am.  $2.81974. 

38.  A  and  B  travelled  on  the  same  road,  and  at  the  same 
rate,  from  Portland  to  Boston.  When  A  was  at  50  miles'  dis- 
tance from  Boston  he  overtook  a  drove  of  geese,  which  were  pro- 
ceeding at  the  rate  of  3  miles  in  2  hours ;  and,  two  hours  after- 
wards, met  a  stage-wagon,  which  was  moving  at  the  rate  of  9 
miles  in  4  hours.  B  overtook  the  same  drove  of  geese  when  he 
was  45  miles  distance  from  Boston,  and  met  the  stage-wagon 
exactly  40  minutes  before  he  arrived  within  31  miles  of  Boston 
Where  was  B  when  A  arrived  at  Boston  ? 

A?ts.  25  miles  from  Boston. 

39.  A  gentleman  has  two  sons,  John  and  Nathan.     John  is 

10  years  old,  and  Nathan  is  15.  He  wishes  to  divide  $1000 
between  his  sons,  in  such  a  manner  that  each,  by  depositing  his 
share  in  a  savings'  bank  which  pays  5  per  cent,  compound  in- 
terest, shall  have  the  same  amount  in  the  bank  when  he  is  21 
years  old.     What  smn  shall  each  deposit  ? 

Am.  John,  $439.30 ;  Nathan,  $560.70. 

40.  My  garden  is  100  feet  square,  and  I  wish  to  raise  its 
surface  2  feet  with  the  soil  taken  from  a  ditch  with  which  I 
intend  to  surround  it.  This  ditch  is  to  be  5  feet  deep,  and  out- 
side the  garden  ;  what  should  be  its  width  ?    Ans.  d.l-\-  feet. 

41.  A  engaged  to  reap  a  field  for  $10,  which  he  would  do  in 
10  days ;  but  after  he  had  labored  2  days  he  engaged  B,  by 
whose  aid  he  supposed  he  could  finish  the  field  in  3  days.  But, 
B  proving  to  be  a  very  inefficient  workman,  A  was  obliged  to  hire 
C  the  last  two  days,  who  proved  to  be  a  superior  laborer ;  the 
field  was  completed  in  5  days.  Now,  if  he  had  not  hired  C, 
and  A  and  B  had  completed  the  work  themselves,  B  would  have 
received  $1.08J-|  in   addition  to   his  services  for  his  3  days' 

29 


338  4         ALGEBIIA. 

labor.     How  long   would  it  have  required  B  and  C,  each,  to 
reap  the  field  ? 

Alls.  B  could  have  reaped  it  in  11^  days,  C  in  SJ-f  days. 

42.  A  man  travelled  105  miles,  and  then  found  that  if  he  had 
not  travelled  so  fast  by  2  miles  an  hour,  he  would  have  been  6 
hours  longer  in  performing  the  same  journey.  How  many  miles 
did  he  go  per  hour  ?  Atis.  7  miles. 

43.  The  difierence  between  the  hypothenuse  and  base  of  a 
right-angled  triangle  is  6  feet,  and  the  difference  between  the 
hypothenuse  and  perpendicular  is  3  feet.  What  are  the  sides 
of  the  triangle  ?  Ans.  15,  12,  and  9  feet. 

44.  In  a  parcel  which  contains  24  coins  of  silver  and  copper, 
each  silver  coin  is  worth  as  many  pence  as  there  are  copper 
coins;  and  each  copper  coin  is  worth  as  many  pence  as  there  are 
silver  coins,  and  the  whole  is  worth  18  shillings.  -  How  many 
are  there  of  each  ?  Ans.  6  of  one,  and  18  of  the  other. 

4^  The  income  of  a  certain  estate  is  to  be  sold  for  a  term  of 
7  years.  A  offers  to  pay  $300  down^  and  $300  at  the  end  of 
each  year;  B  ofi"ers  $800  doiDn,  and  $250  at  the  end  of  each 
year;_C  offers  $1300  doion^  and  $200  at  the  end  of  each  year; 
D  will  pay  $2500  "  cash  down.''''  Which  has  made  the  best  offer, 
if  interest  is  to  be  reckoned  at  6  per  cent,  compound  interest  ? 

/  Value  of  A's  offer,  $1974.71.4. 

)b'^ 


Ans.    \  B's  offer,  $2195.59.5  ;  C's  offer,  $2416.47.6. 
's  offer,  $2500.     Hence  D's  offer  is  the  best. 


46.  A  gentleman  being  asked  the  age  of  his  two  sons,  replied, 
that  if  the  sum  of  their  ages  were  multiplied  by  the  age  of  the 
elder,  the  product  would  be  144 ;  but  if  the  difference  of  their 
ages  were  multiplied  by  that  of  the  younger,  the  product  would 
be  14.    What  was  the  age  of  each  ?  Ans.  9  and  7. 

47.  The  sum  of  two  numbers  is  20,  and  the  sum  of  their 
cubes  is  2060.     What  are  the  numbers  ?  Ans.  9  and  11. 

48.  If  the  product  of  two  numbers  be  added  to  the  square  of 
the  larger,  the  sum  will  be  112 ;   but,  if  the  square  of  the  less 


MISCELLANEOUS     QUESTIONS.  339 

be  taken  from  their  product,  the  remainder  will  be  12,     Re- 
quired the  numbers.  Ans.  8  and  6. 

49.  AVhat  number  is  that  which,  being  added  to  twice  its 
square  root,  equals  24  ?  Ans.  16. 

50.  If  a  man  owe  $2000,  what  sum  shall  he  pay  daily,  so  as 
to  cancel  the  debt,  principal  and  interest,  at  the  end  of  the 
year,  reckoning  the  interest  at  6  per  cent,  ?        Ans.  $5.0394. 

51.  I  have  84i-  square  feet  of  plank,  that  is  3  inches  thick.  T 
How  large  a  cubical  box  can  be  made  from  it  ? 

Alts.  Each  side  measures  48  inches. 

52.  From  62f|-  feet  of  plank,  that  is  2h  inches  thick,  I  wish 
to  make  a  box  whose  length  shall  be  four  times  its  width,  and 
whose  height  and  width  shall  be  equal.     What  are  its  dimen-     ' 
sions^  Ans.  Length  8  feet,  width  and  height  2  feet. 

5&.  There  was  a  cask  containing  20  gallons  of  wine ;  a  cer- 
tain quantity  of  this  was  drawn  off  into  another  cask  of  equal 
size,  and  this  last  filled  with  water,  and  afterwards  the  first  cask 
was  filled  with  the  mixture.  It  now  appears  that,  if  6f  gallons 
of  the  mixture  be  drawn  off  from  the  first  into  the  second  cask, 
there  will  be  equal  quantities  of  wine  in  each.  What  was  the 
quantity  of  wine  drawn  off  at  first?  Ans.  10  gallons.. 

54.  Aftor  A  had  travelled  for  2f  hours,  at  the  rate  of  4  miles 
an  hour,  B  set  out  to  overtake  him ;  and,  in  order  thereto,  went 
four  m\[Q§  and  a  half  the  first  hour,  four  and  three-quarters  the 
second,  five  the  third,  and  so  on,  gaining  a  quarter  of  a  mile 
every  hour.     In  how  many  hours  would  he  overtake  A  ? 

Alls.  8  hours. 

55.  The  sum  of  the  first  and  second  of  four  numbers  in  geo- 
metrical progression  is  15,  and  the  sum  of  the  third  and  fourth 
is  60.     Required  the  numbers.  Ans.  5,  10,  20,  40. 

56.  The  sum  of  the  squares  of  the  extremes  of  four  numbers 
in  arithmetical  progression  is  200,  and  the  sum  of  the  squares 
of  the  means  is  136.     What  are  the  numbers  ? 

^^:  Am.  14,  10,  6,2. 


■^^SSO  ALGEBRA."  ^/''^'^^ 

57.  A  tailor  bought  a  piece  of  cloth  for  £147,  from  which  he 
cut  off  12  yards  for  his  own  use ;  he  sold  the  remainder  for 
£120  5^.,  gaining  5  shillings  per  yard.  How  many  yards  were 
there,  and  what  did  it  cost  him  per  yard  ? 

A71S.  49  yards,  at  £3  per  yard. 

58.  In  a  mixture  of  rye  and  wheat,  the  difference  between 
the  quantities  of  each  is  to  the  quantity  of  wheat  as  100  is  to 
the  number  of  bushels  of  rye,  and  the  same  difference  is  to  the 
quantity  of  rye  as  4  to  the  number  of  bushels  of  wheat.  How 
many  bushels  are  there  of  each  ? 

A71S.  25  bushels  of  rye,  and  5  of  wheat. 

59.  It  is  required  to  find  two  numbers,  such  that  the  product 
of  the  greater  into  the  square  root  of  the  less  shall  be  equal  to 
48,  and  the  product  of  the  less  into  the  square  root  of  the 
greater  may  be  36.  A71S.  16  and  9. 

60.  If  the  difference  of  two  numbers  be  multiplied  by  the 
greater,  and  the  product  divided  by  the  less,  the  result  will  be 
48;  but,  if  the  difference  be  multiplied  by  the  less,  and  the 
product  divided  by  the  greater,  the  result  will  be  3.  What  are 
the  numbers  ?  Atis.  16  and  4. 

61.  Find  two  numbers,  such  that  the  square  of  the  greater, 
multiplied  by  the  less,  shall  be  equal  to  448;  and  the  square 
of  the  less,  multiplied  by  the  greater,  shall  be  392. 

Atis.    8  and  7. 

62.  If  two  numbers  be  each  multiplied  by  27,  the  first  pro- 
duct is  a  square,  and  the  second  the  square  root  of  that  square; 

^^'    but,  if  each  be  multiplied  by  3,  the  first  product  is  a  cube,  and 
the  second  the  cube  root  of  that  cube.     What  are  the  numbers  ? 
ji  7  Atis.  243  and  3. 

'^  63.  A  farmer  has  two  cubical  stacks  of  hay ;  the  side  of  one  is 
3  yards  longer  than  the  side  of  the  other,  and  the  difference  of 
their  contents  is  117  solid  yards.     Kequired  the  side  of  each. 

Atis.  5  and  2  yards. 

64.  A  gentleman  started  from  Boston  for  New  York ;  he  trav- 
elled 20  miles  the  first  day,  18  miles  the  second  day,  and  16 


MISCELLANEOUS     QUESTIONS.  341 

miles  the  third  day,  so  continuing  to  travel  two  miles  less  each 
day  than  the  former.  How  far  was  the  gentleman  from  Boston 
at  the  end  of  the  twentieth  day  ?  A7is. 

65.  A  certain  farm  is  a  parallelogram,  and  a  diagonal  line 
from  one  corner  to  the  opposite  is  60  rods,  and  the  longer  side 
is  to  the  shorter  as  4  to  3.     Required  the  contents  of  the  farm. 

Ans.  10  Acres,  3  Hoods,  8  Poles. 

66.' A  gentleman  asking  a  lady  her  age,  she  replied.  If  you 
add  the  square  root  of  it  to  half  of  it,  and  subtract  12,  there  will 
remain  nothing.     Required  her  age.  A^is.  16. 

67.  What  number  is  that  to  which  if  1,  7,  and  19  be  sever- 
ally added,  the  first  sum  shall  have  the  same  ratio  to  the  second 
that  the  second  has  to  the  third  ?  Atis.  5. 

68.  The  sum  of  two  numbers  is  12,  and  they  have  the  same 
ratio  to  each  other  that  their  difierence  has  to  40.  AVhat  are 
the  numbers  ?  Ans.  2  and  10. 

69.  There  are  two  numbers  whose  product  is  54,  and  the 
greater  is  to  the  less  as  their  sum  is  to  10.  What  are  those 
numbers  ?  A7is.  9  and  6. 

«^  70.  Divide  20  into  two  such  parts  that  the  square  of  the 
greater  shall  be  to  the  square  of  the  less  as  9  to  4.  What  are 
those  parts  ?  A?is.  12  and  8. 

"  71.  Let  24  be  divided  into  two  such  parts  that  the  quotient 
of  the  greater  divided  by  the  less  shall  be  to  the  quotient  of  the 
less  divided  by  the  greater  as  9  to  1.  A?is.  18  and  6. 

72.  Divide  14  into  two  such  parts  that  their  squares  shall  be 
to  each  other  as  9  to  16.  Am.  6  and  8. 

6  73.  Divide  12  into  two  such  parts  that  the  sum  of  their 
squares  shall  be  to  the  difference  of  their  squares  as  5  to  3. 

Ans.  4  and  8. 

74.  There  are  two  numbers,  whose  product  is  12,  and  the 
sum  of  whose  cubes  is  to  the  cube  of  their  sum  as  91  to  343. 
What  are  the  numbers  ?  Ayis.  3  and  4. 

29^ 


342  ALGEBRA. 

75.  The  product  of  two  numbers  is  120 ;  and,  if  the  greater 
be  increased  by  8  and  the  less  by  5,  the  product  of  the  two  num- 
bers will  be  300.     What  are  the  numbers  ?     Atis.  10  and  12. 

-  76.  A,  B,  and  C,  make  a  joint  stock ;  A  puts  in  $60  less 
than  B,  and  $68  more  than  C,  and  the  sum  of  the  shares  of 
A  and  B  is  to  the  sum  of  the  shares  of  B  and  C  as  5  to  4. 
What  did  each  put  in  ? 

\  ;  -«j   -^   -;  ,      A71S.  A  put  in  $140,  B  $200,  and  C  $72. 

77.  A  and  B  engage  in  speculation,  with  different  sums ;  A 
gains  $150,  B  loses  $50.  Now  A's  stock  is  to  B's  as  3  to  2 ; 
but,  had  A  lost  $50,  and  B  gained  $100,^then  A's  stock  would 
have  been  to  B's  as  5  to  9.     What  was  the  stock  of  each  ? 

Ans.  A's,  $300 ;  B's,  $350. 

78.  Find  two  numbers  in  the  ratio  of  5  to  7,  to  which  two 
other  numbers,  in  the  ratio  of  3  to  5,  being  respectively  added, 
the  sums  shall  be  in  the  ratio  of  9  to  13,  and  the  difference  of 
the  sums  shall  be  16.  (  First  two  numbers,  30  and  42. 

(  Last  two  numbers,  6  and  10. 

79.  A  merchant  mixes  wheat,  which  costs  10  shillings  per 
bushel,  with  barley,  which  costs  him  4  shillings  per  bushel,  in 
such  proportion  as  to  gain  43|-  per  cent.,  by  selling  the  mixture 
at  11  shillings  per  bushel.     Required  the  proportion. 

Ans.  He  must  mix  14  bushels  of  wheat  with  9  of  barley. 

80.  A  and  B  can  dig  a  cellar  in  a  days,  A  and  C  can  do  the 

labor  in  b  days,  and  B  and  C  can  do  the  same  in  c  days.     In 

what  time  would  each  perform  the  labor,  and  how  long  would  it 

require  A,  B,  and  C,  to  complete  the  work  ? 

.         .    .  2abc        ^         _  .  2abc  ,  „     . 

A?is.  A  in  — — days,  B  in     ,  .  , days,     C     in 

ac-^bc — ab  ab-\-bc — ac 

days,  and  A,  B,  C,  in  — ~  days. 


ab-\-ac—bc      *^  '  '     '     '       ^^_j_^c4-^c 

81.  A  and  B  made  a  joint  stock  of  $833,  which,  after  a  suc- 
cessful speculation,  produced  a  clear  gain  of  $153.     Of  this  B 
had  $45  more  than  A.     What  did  each  person  contribute  to  the 
,.,  stock?  ^7w.  B  $539,  and  A  $294. 


MISCELLANEOUS     QUESTIONS.  343 

82.  A  gentleman  having  asked  a  ladj  her  age,  she  modestly 
replied,  that  if  she  were  four  years  younger,  and  he  were  four 
years  older,  his  age  would  be  twice  that  of  hers ;  but,  if  she  were 
four  years  older,  and  he  were  four  years  younger,  their  ages 
would  be  the  same.     What  was  the  age  of  each  ? 

Ans.  Gentleman's  age,  28  years  ;  lady's  age,  20  years. 

ALGEBRA   APPLIED   TO    GEOMETRY. 

83.  Suppose  a  tree,  48  feet  in  height,  to  stand  on  a  hori- 
zontal plane.  At  what  height  from  the  ground  must  it  be  cut 
off,  so  that  the  top  of  it  may  fall  on  a  point  24  feet  from  the 
bottom  of  the  tree,  the  end,  where  it  was  cut  off,  resting  on  the 
stump  ?  Ans.  18  feet. 

84.  A  certain  man,  owning  a  farm  lying  in  a  circle,  gave  it 
in  his  will  to  his  wife,  four  sons,  and  four  daughters,  as  follows : 
to  his  sons  he  gave  four  circles,  as  large  as  could  be  drawn 
within  the  circumference  of  the  farm ;  to  his  daughters  he  gave 
the  four  spaces  lying  between  the  son's  circles  and  the  circum- 
ference of  the  farm,  and  to  his  wife  he  gave  the  part  remaining 
in  the  centre,  which  contained  just  one  acre.  How  much  did 
the  whole  farm  contain,  how  much  did  each  son  have,  and  how 
much  did  each  daughter  have  ? 

t  The  farm  contained  21  Acres,  1  Kood,  12    Poles. 

Ans.  }  Each  son  had  3  Acres,  2  Roods,  25^-  Poles. 

(  Each  daughter  had     1  Acre,  1  Rood,  274-  Poles. 

85.  A  gentleman  has  a  garden  in  the  form  of  an  equilateral 
triangle,  the  sides  whereof  are  each  100  feet.  At  each  corner  of 
the  garden  stands  a  tower;  the  height  of  the  first  tower  is  40 
feet,  that  of  the  second  45  feet,  and  that  of  the  third  is  55  feet. 
At  what  distance  from  the  bottom  of  each  of  these  towers  must  a 
ladder  be  placed,  that  it  may  just  reach  the  top  of  each  tower ; 
and  what  must  be  the  length  of  the  ladder,  the  ground  of  the 
garden  being  horizontal  ? 

Ans.  From  the  foot  of  the  ladder  to  the  base  of  the  first 
tower,  63.273+  feet;  second  tower,  59.820-f-  feet;  third  tower, 
50.779+  feet;  length  of  the  ladder,  74.856+  feet. 


344  ALGEBRA. 

86.  If  c  be  tlie  hjpotlienuse  of  a  rigbt-angled  triangle,  b  the 
base,  and  a  the  perpendicular,  it  is  required  to  find  the  segments 
made  by  a  perpendicular  drawn  from  the  right  angle  to  the 
hypothenuse.  .         b'^-\-c' — a?       ^  a?-\-c^ — b'^ 

'Ic       ^  2c       * 

87.  From  a  point  within  an  equilateral  triangle,  there  are 
drawn  three  perpendiculars  to  the  several  sides ;  the  length  of 
the  first  is  20  feet,  the  second  30  feet,  and  the  third  36  feet. 
Kequired  the  length  of  the  sides  of  the  triangle. 

Ans.  49.652-f  feet. 

88.  A  sphere  of  gold,  whose  diameter  is  one  inch,  weighs  10 
ounces,  and  each  ounce  is  valued  at  $16.  What  is  the  value 
of  5  spheres  of  gold,  whose  several  diameters  are  1,  2,  3,  4, 
and  5  inches?  Am.  $3600. 

89.  There  is  a  loaf  of  bread,  which  is  half  a  sphere,  whose 
diameter  measures  12  inches.  How  thick  must  the  crust  be 
baked,  that  the  remainder  shall  be  half  the  contents  of  the  loaf? 

Ans.  .8038-1-  inch. 

90.  There  are  two  towers  of  unequal  heights,  situated  on  a 
plane,  near  each  other.  A  line  extending  from  the  base  of  the 
less  to  the  top  of  the  larger  is  100  feet ;  and  a  line  from  the 
base  of  the  larger  to  the  top  of  the  less  is  80.27-{-  feet;  a  per- 
pendicular let  fall  from  the  point  where  the  lines  cross  each 
other,  to  the  surface  of  the  plane,  is  32  feet.  Required  the 
height  of  the  towers,  and  their  distance  from  each  other. 

(  Height  of  the  larger  tower,  80  feet. 
Ans.  }  Height  of  the  less,  53^  feet. 

(  Distance  between  the  towers,  60  feet. 

91.  There  is  a  conical  glass,  6  inches  deep  ;  the  diameter  at 
the  top  is  5  inches,  and  it  is  \  full  of  water.  If  a  ball  4  inches 
in  diameter  be  put  into  this  glass,  how  much  of  its  axis  will 
be  immersed  in  the  water  ?  Ans.  .546  inch. 

92.  How  many  balls  1  inch  in  diameter  can  be  put  into  a 
cubical  box  whose  sides  measure  each  one  foot  in  the  dear  ? 

Ans.  2151  balls. 


TABLE, 


CONTAINING  THE 


LOGARITHMS   OF   NUMBERS 


FROM  1   TO   10,000. 


Numbers  from  1  to  100  and  their  Logarithms,  with  their  Indices. 


No. 

Log. 

No. 

Log. 

No. 

Log. 

No. 

Log. 

No. 

Log. 

1 

0.000000 

21 

1.322219 

41 

1.612784 

61 

1.785330 

81 

1.908485 

2 

0.301030 

22 

1.342423 

42 

1.623249 

62 

1.792392 

82 

1.913814 

3 

0.477121 

23 

1.361728 

43 

1.633468 

63 

1.799341 

83 

1.919078 

4 

0.G020G0 

24 

1.380211 

44 

1.643453 

64 

1.806180 

84 

1.924279 

5 

0.698970 

25 

1.397940 

45 

1.653213 

65 

1.812913 

85 

1.929419 

6 

0.778151 

2G 

1.414973 

46 

1.662758 

66 

1S19544 

86 

1.934498 

7 

0.845098 

27 

1.431364 

47 

1.672098 

67 

1.82G075 

87 

1.939519 

8 

0.903090 

28 

1.447158 

48 

1.681241 

68 

1.832509 

88 

1.944483 

9 

0.954243 

29 

1.462398 

49 

1.690196 

69 

1.838849 

89 

1.949390 

10 

1.000000 

30 

1.477121 

50 

1.698970 

70 

1.845098 

90 

1.954243 

11 

1.041393 

31 

1.491362 

51 

1.707570 

71 

1.851258 

91 

1.959041 

12 

1.079181 

32 

1.505150 

52 

1.716003 

72 

1.857332 

92 

1.9G3788 

13 

1.113943 

33 

1.518514 

53 

1.72427G 

73 

1.863323 

93 

1.9G8483 

14 

1.14G128 

34 

1.531479 

54 

1.732394 

74 

1.869232 

94 

1.973128 

15 

1.17G091 

35 

1.544068 

55 

1.7403G3 

75 

1.875061 

95 

1.977724 

IG 

1.204120 

36 

1.556303 

56 

1.748188 

76 

1.880814 

96 

1.982271 

17 

1.230449 

37 

1.568202 

i)i 

1.755875 

77 

1.886491 

97 

1.986772 

18 

1.255273 

38 

1.579784 

58 

1.763428 

78 

1.892095 

98 

1.99122C 

19 

1.278754 

39 

1.591065 

59 

1.770852 

79 

1.897627 

99 

1.995G35 

20 

1.301030 

40 

1.602060 

00 

1.778151 

80 

1.903090 

100 

2.000000 

Note.  —  In  the  following  part  of  the  Table  the  Indices  are  omitted,  as  they 
can  be  very  easily  supplied  by  the  directions  given  in  Section  xxxs.,  p.  270,  on 

Logarithms. 


346 


LOGARITHMS 


N. 

0 

1   1   2 

3 

4. 

5   1   6 

7 

8 

9   |D.| 

100 

000000 

000434 

0008G8 

001301 

001734 

0021G6 

002598 

003029 

003461 

003891 

432 

1 

4321 

4751 

5181 

5609 

6038 

6466 

6894 

7321 

7748 

8174 

428 

2 

8G00 

9026 

9451 

9876 

010300 

010724 

011147 

011570 

011993 

012415 

424 

3 

012837 

013259 

013680 

014100 

4521 

4940 

5360 

5779 

6197 

6616 

420 

4 

7033 

7451 

7868 

8284 

8700 

9116 

9532 

9947 

020361 

020775 

416 

5 

021189 

021G03 

022016 

022428 

022841 

023252 

023664 

024075 

4486 

4896 

412 

i  ^ 

5306 

5715 

G125 

G533 

6942 

7350 

7757 

8164 

8571 

8978 

408 

7 

9384 

9789 

030195 

030600 

031004 

031408 

031812 

032210  032619 

033021 

404 

8 

033424 

033826 

4227 

4628 

5029 

5430 

5830 

6230   6629 

7028 

400 

i  9 

742G 

7825 

8223 

8620 

9017 

9414 

9811 

040207  040602 

040998 

397 

|110  041393 

041787 

042182 

042576 

042969 

043362 

043755 

044148 

044540 

044932 

393 

1   5323 

5714 

6105 

6495 

6885 

7275 

7664 

8053 

8442 

8830 

390 

2   9218 

9606 

9993 

050380 

050766 

051153 

051538 

051924 

052309 

052694 

386 

3 

053078  0534G3 

053846 

4230 

4613 

4996 

5378 

5760 

6142 

6524 

383 

4 

G905   7286 

7666 

8046 

8426 

8805 

9185 

9563 

9942 

060320 

379 

5 

060698  0C1075 

061452 

061829 

062206 

062582 

062958 

063333 

063709 

4083 

376 

6 

4458 

4832 

5206 

5580 

5953 

6326 

6699 

7071 

7443 

7815 

373 

7 

8186 

8557 

8928 

9298 

9668 

070038 

070407 

070776 

071145071514 

370 

8 

071882 

072250 

072617 

072985 

073352 

3718 

4085 

4451 

4816   5182 

366 

9 

5547 

5912 

627G 

6640 

7004 

7368 

7731 

8094 

8457 

8819 

363 

120 

079181 

079543 

079904 

080266  080626 

080987 

081347 

081707 

082067 

082426 

360 

1 

082785 

083144 

083503 

3861 

4219 

4576 

4934 

5291 

5647 

G004 

357 

2 

6360 

6716 

7071 

7426 

7781 

8136 

8490 

8845 

9198 

9552 

355 

3 

9905 

090258 

090611 

090963 

091315 

091667 

092018 

092370 

092721 

093071 

352 

!  4 

093422 

3772 

4122 

4471 

4820 

5169 

5518 

5866 

6215 

6562  349 

1  5 

6910 

7257 

7604 

7951 

8298 

8644 

8990 

9335 

9681 

100026  346 

6 

100371 

100715 

101059 

101403 

101747 

102091 

102434 

102777 

103119 

3462  343 

i  "^ 

3804 

4146 

4487 

4828 

5169 

5510 

5851 

6191 

6531 

6871 341 

8 

7210 

7549 

7888 

8227 

8565 

8903 

9241 

9579 

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LOGARITHMS 


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0 

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340 

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2 

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D. 

OF    NUMBERS. 


351 


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7   1   «   1   9   1  1). 

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2 

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6 

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100 

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106 

2 

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5213 

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3 

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8360 

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96 

1 

4177 

4273 

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4465 

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4754 

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2 

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7056 

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5 

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95 

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1813 

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2569   2663 

95 

IT 

0 

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352 


LOGARITHMS 


N. 

0 

1 

2 

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4 

5   1   6 

7   1   8 

9 

D. 

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2 

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2568 

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68 

5 

2774 

2842 

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3116 

3184 

3252 

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68 

6 

3457 

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68 

7 

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68 

8 

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D. 

OF    NUMBERS. 


355 


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4 

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67 

5 

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9627 

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67 

0 

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67 

7 

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67 

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67 

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1 

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67 

2 

4248 

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67 

3 

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66 

4 

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66 

5 

6241 

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66 

6 

6904 

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7169 

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66 

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660 

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4 

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8   1230 

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9   1870   1934 

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356 


LOGARITHMS 


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0 

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6 

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61 

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61 

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61 

6 

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61 

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5519 

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61 

8 

6124 

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OF    NUMBERS. 


357' 


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4174 

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4283 

4337 

4391 

4445 

4499 

4553 

4607 

4661 

54 

3 

4716 

4770 

4824 

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4932 

4986 

6040 

6094 

5148 

5202 

54 

4 

5256 

5310 

5364 

6418 

6472 

6526 

5580 

6634 

5688 

5742 

54 

5 

6796 

5850 

6904 

6958 

0012 

6066 

6119 

6173 

6227 

6281 

54 

6 

6335 

0389 

6443 

6497 

6551 

6604 

6658 

6712 

6766 

6820 

54 

7 

6874 

6927 

6981 

7035 

7089 

7143 

7196 

7250 

7304 

7358 

54 

8 

7411 

7465 

7519 

7573 

7626 

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7734 

7787 

7841 

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54 

9 

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8002 

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54 

810,908485 

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54 

1 

9021 

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9128 

9181 

9235 

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9342 

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9449   9503 

54 

2 

9556 

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9716 

9770 

9823 

9877 

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9984910037 

53 

31910091 

910144 

910197 

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910304 

910358  910411 

910464 

910518!  0571 

53 

4 

0624 

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0731 

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0838 

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0944 

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1051 

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53 

6 

1158 

1211 

1264 

1317 

1371 

1424 

1477 

1530 

1584 

1637 

53 

G 

1690 

1743 

1797 

1850 

1903 

1956 

2009 

2063 

2116 

2169 

53 

7 

2222 

2275 

2328 

2381 

2435 

2488 

2541 

2594 

2647 

2700 

53 

8 

2753 

2806 

2859 

2913 

2966 

3019 

3072 

3125 

3178 

3231 

53 

9 

3284 

3337 

3390 

3443 

3496 

3549 

3002 

3655 

3708 

3761 

53 

N. 

0     1   1   2   1   3   1   4   1 

5     6   1   7   1   8   1 

'9   1 

D. 

358 


LOGARITHMS 


N.l   0  1 

1  1   2 

3 

4   1 

5   1   6 

7 

8   1   9   |D.| 

820 

913814 

913867 

913920 

913973 

914026 

914079 

914132 

914184 

914237 

914290 

53 

1 

4343 

4396 

4449 

4502 

4555 

4608 

4660 

4713 

4766 

4819 

53 

2 

4872 

4925 

4977 

5030 

6083 

5136 

6189 

5241 

5294 

6347 

53 

3 

5400 

5453 

5505 

6558 

5611 

6664 

5716 

6769 

6822 

6875 

53 

4 

6927 

5980 

6033 

6085 

6138 

6191 

6243 

6296 

6349 

6401 

53 

5 

6454 

6507 

6559 

6612 

6664 

6717 

6770 

6822 

6875 

6927 

53 

6 

6980 

7033 

7085 

7138 

7190 

7243 

7295 

7348 

7400 

7453 

53 

7 

7506 

7558 

7611 

7663 

7716 

7768 

7820 

7873 

7925 

7978 

52 

8 

8030 

8083 

8135 

8188 

8240 

8293 

8345 

8397 

8450 

8502 

62 

9 

8555 

8607 

8659 

8712 

8764 

8816 

8869 

8921 

8973 

9026 

52 

830  919078 

919130 

919183 

919235 

919287 

919340 

919392 

919444 

919496 

919549 

52 

1 

9G01 

9653 

9706 

9758 

9810 

9862 

9914 

9967 

920019 

920071 

52 

2 

920123 

920176 

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920280 

920332 

920384 

920436 

920489 

0541 

0593 

52 

3 

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0749 

0801 

0853 

0906 

0958 

1010 

1062 

1114 

52 

4 

1166 

1218 

1270 

1322 

1374 

1426 

1478 

1530 

1582 

1634 

52 

5 

1686 

1738 

1790 

1842 

1894 

1946 

1998 

2050 

2102 

2154 

52 

G 

2206 

2258 

2310 

2362 

2414 

2466 

2518 

2570 

2622 

2674 

52 

7 

2725 

2777 

2829 

2881 

2933 

2985 

3037 

3089 

3140 

3192 

52 

8 

3244 

3296 

3348 

3399 

3451 

3503 

3555 

3607 

3658 

3710 

62 

9 

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3814 

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52 

840 

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52 

1 

4796 

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5003 

6054 

6106 

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5209 

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62 

2 

5312 

5364 

5415 

5467 

5518 

6570 

6621 

5673 

5725 

6776 

62 

3 

5828 

5879 

6931 

5982 

6034 

6085 

6137 

6188 

6240 

6291 

51 

4 

6342 

6394 

6445 

6497 

6548 

6600 

6651 

6702 

6754 

6805 

51 

5 

6857 

6908 

6959 

7011 

7062 

7114 

7165 

7216 

7268 

7319 

51 

6 

7370 

7422 

7473 

7524 

7576 

7627 

7678 

7730 

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7832 

61 

7 

7883 

7935 

7986 

8037 

8088 

8140 

8191 

8242 

8293 

8345 

51 

8 

8396 

8447 

8498 

8549 

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8703 

8754 

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8857 

51 

9 

8908 

8959 

9010 

9061 

9112 

9163 

9215 

9266 

9317 

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850 

929419 

929470 

929521 

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51 

1 

9930 

9981 

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930083 

930134 

930185 

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930287 

930338 

930389 

51 

2 

930440 

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0542 

0592 

0643 

0694 

0745 

0796 

0847 

0898 

51 

3 

0949 

1000 

1051 

1102 

1153 

1204 

1254 

1305 

1356 

1407 

51 

4 

1458 

1509 

1560 

1610 

1661 

1712 

1763 

1814 

1865 

1915 

51 

5 

1966 

2017 

2068 

2118 

2169 

2220 

2271 

2322 

2372 

2423 

51 

6 

2474 

2524 

2575 

2626 

2677 

2727 

2778 

2829 

2879 

2930 

51 

7 

2981 

3031 

3082 

3133 

3183 

3234 

3285 

3335 

3386 

3437 

51 

8 

3487 

3538 

3589 

3639 

3690 

3740 

3791 

3841 

3892 

3943 

51 

9 

3993 

4044 

4094 

4145 

4195 

4246 

4296 

4347 

4397 

4448 

51 

860 

934498 

934549 

934599 

934650 

934700 

934751 

934801 

934852 

934902 

934953 

60 

1 

5003 

5054 

6104 

6154 

6205 

5255 

6306 

6356 

6406 

5457 

50 

2 

5507 

5558 

5608 

5658 

5709 

6759 

5809 

6860 

6910 

6960 

60 

3 

6011 

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6111 

6162 

6212 

6262 

6313 

6363 

6413 

6463 

50 

4 

6514 

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G614 

6665 

6715 

6765 

6816 

6865 

6916 

6966 

50 

5 

7016 

7066 

7117 

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7217 

7267 

7317 

7367 

7418 

7468 

50 

G 

7518 

7568 

7618 

7668 

7718 

7769 

7819 

7869 

7919 

7969 

50 

7 

8019 

8069 

8119 

8169 

8219 

8269 

8320 

8370 

8420 

8470 

60 

8 

8520 

8570 

8620 

8670 

8720 

8770 

8820 

8870 

8920 

8970 

60 

9 

9020 

9070 

9120 

9170 

9220 

9270 

9320 

9369 

9419 

9469 

50 

870 

939519 

969569 

939619 

939669 

939719 

939769 

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939869 

939918 

939968 

60 

1 

940018 

940068 

940118 

940168 

940218 

940267 

940317 

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940417 

940467 

50 

2 

0516 

056G 

0616 

0666 

0716 

0765 

0816 

0865 

0915 

0964 

50 

3 

1014 

1064 

1114 

1163 

1213 

1263 

1313 

1362 

1412 

1462 

50 

4 

1511 

1561 

1611 

1G60 

1710 

1760 

1809 

1859 

1909 

1958 

50 

5 

2008 

2058 

2107 

2157 

2207 

2256 

2306 

2355 

2405 

2455 

50 

6 

2504 

2554 

2603 

2653 

2702 

2752 

2801 

2851 

2901 

2950 

50 

7 

3000 

3049 

3099 

3148 

3198 

3247 

3297 

3346 

3396 

3446 

49 

8 

3445 

3544 

3593 

3643 

3692 

3742 

3791 

3841 

3890 

3939 

49 

9 

3989 

4038 

4088 

4137 

4186 

423G 

4285 

4335 

4384 

4433 

49 

N. 

0 

1   1   2 

3 

4 

1  6 

6 

7 

8 

9 

D. 

OF    NUMBERS. 


359 


N.|   0 

i 

2 

3 

4   1   .-, 

.;  1  7  1   b 

9   ,I».| 

am 

944483 

944532 

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49 

1 

4976 

6025 

5074 

5124 

5173 

5222 

5272 

5321 

5370 

5419 

49 

2 

5469 

5518 

5567 

5616 

5665 

5715 

5764 

5813 

5862 

5912 

49 

3 

5961 

6010 

6059 

6108 

6157 

6207 

6250 

6305 

6354 

6403 

49 

4 

6452 

6501 

6551 

6600 

6649 

6698 

6747 

6796 

6845 

6894 

49 

5 

6943 

0992 

7041 

7090 

7140 

7189 

7238 

7287 

7336 

7385 

49 

6 

7434 

7483 

7532 

7581 

7630 

7679 

7728 

7777 

7826 

7875 

49 

7 

7924 

7973 

8022 

8070 

8119 

8168 

8217 

8266 

8315 

8364 

49 

8 

8413 

8462 

8511 

8560 

8609 

8657 

8706 

8755 

8804 

8853 

49 

9 

8902 

8951 

8999 

9048 

9097 

9146 

9195 

9244 

9292 

9341 

49 
l9 

890 

949390 

949439 

949488 

949536 

949585 

949634 

949683 

949731 

9497801949829 

1 

9878 

9926 

9975 

950024 

950073, 

950121 

950170 

950219 

950267 

950310 

49 

2 

950365 

950414 

950462 

0511 

0560 

0608 

0657 

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0754 

0803 

49 

3 

0851 

0900 

0949 

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1046 

1095 

1143 

1192 

1240 

1289 

49 

4 

1338 

1386 

1435 

1483 

1532 

1580 

1629 

1677 

1726 

1775 

49 

5 

1823 

1872 

1920 

1969 

2017 

2066 

2114 

2163 

2211 

2260 

48 

6 

2308 

2356 

2405 

2453 

2502 

2550 

2599 

2647 

2696 

2744 

48 

7 

2792 

2841 

2889 

2938 

2986 

3034 

3083 

3131 

3180 

3228 

48 

8 

3276 

3325 

3373 

3421 

3470 

3518 

3566 

3615 

3663 

3711 

48 

9 

3760 

3808 

3856 

3905 

3953 

4001 

4049 

4098 

4146 

4194 

48 

900  954243 

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48 

1 

4725 

4773 

4821 

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4918 

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5014 

5062 

5110 

5158 

48 

2 

5207 

5255 

6303 

5351 

5399 

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5495 

5543 

5592 

5640 

48 

3 

5688 

5736 

6784 

5832 

6880 

6928 

6976 

6024 

6072 

6120 

48 

4 

6168 

6216 

6265 

6313 

6361 

6409 

6457 

6505 

6553 

6601 

48 

6 

6649 

6697 

6745 

6793 

6840 

6888 

6936 

6984 

7032 

7080 

48 

6 

7128 

7176 

7224 

7272 

7320 

7368 

7416 

7464 

7512 

7559 

48 

7 

7607 

7655 

7703 

7751 

7799 

7847 

7894 

7942 

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8038 

48 

8 

8086 

8134 

8181 

8229 

8277 

8325 

8373 

8421 

8468 

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48 

9 

8564 

8612 

8659 

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8850 

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48 

910 

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48 

1 

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48 

2 

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48 

3  960471 

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0804 

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48 

4 

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1041 

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1136 

1184 

1231 

1279 

1326 

1374 

48 

5 

1421 

1469 

1516 

1563 

1611 

1658 

1706 

1753 

1801 

1848 

47 

6 

1895 

1943 

1990 

2038 

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2132 

2180 

2227 

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47 

7 

2369 

2417 

2464 

2511 

2559 

2606 

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2701 

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47 

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2843 

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2937 

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3126 

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47 

d'     3316 

3363 

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3457 

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47 

9201963788 

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964118 

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47 

1 

4260 

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4401 

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4495 

4542 

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47 

2 

4731 

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4825 

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4919 

4966 

5013 

5061 

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47 

3 

5202 

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5296 

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5390 

5437 

5484 

6531 

5578 

5625 

47 

4 

5672 

5719 

5766 

5813 

6860 

5907 

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6001 

6048 

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47 

5 

0142 

6189 

6236 

6283 

6329 

6376 

6423 

6470 

6517 

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47 

6 

6011 

6658 

6705 

6752 

6799 

6845 

6892 

6939 

6986 

7033 

47 

7 

7080 

7127 

7173 

7220 

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7314 

7361 

7408 

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7501 

47 

8 

7548 

7595 

7642 

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7829 

7875 

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47 

9 

8016 

8062 

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8156 

8203 

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47 

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968530 

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968716 

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968810 

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47 

1 

8950 

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9043 

9090 

9136 

9183 

9229 

9276 

9323 

9369 

47 

2 

9416 

9463 

9509 

9556 

9602 

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9742 

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9835 

47 

3 

9882 

9928 

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970068 

970114 

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47 

4 

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970440 

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46 

5 

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0904 

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1044 

1090 

1137 

1183 

1229 

46 

C 

1276 

1322 

1369 

1415 

1461 

1508;  1554 

1601 

1647 

1693 

46 

7 

1740 

1786 

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1925 

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2064 

2110 

2157 

46 

8 

2203 

2249 

2295 

2342 

2388 

2434 

2481 

2527 

2573 

2619 

46 

9 

2666 

2712 

2758 

2804 

2851 

2897 

2943 

2989 

3035 

30S2|  46 

N.|   0 

1   1   2   1   3   1   4 

II   5 

6   1   7   1   8 

9   1  D. 

860 


LOGARITHMS    OF    NUMBERS. 


N.|   0   ' 

1  1   2   1   3   1 

4   1 

5 

6 

7 

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973451 

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46 

1 

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3636 

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3728 

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3913 

3959 

4005 

46 

2 

4051 

4097 

4143 

4189 

4235 

4281 

4327 

4374 

4420 

4466 

46 

3 

4512 

4558 

4604 

4650 

4696 

4742 

4788 

4834 

4880 

4926 

46 

4 

4972 

5018 

5064 

5110 

5156 

5202 

5248 

5294 

5340 

5386 

46 

5 

5432 

5478 

5524 

5570 

5616 

5662 

5707 

5753 

5799 

5845 

46 

6 

5891 

5937 

5983 

6029 

6075 

6121 

6167 

6212 

6258 

6304 

46 

7 

6350 

6396 

6442 

6488 

6533 

6579 

6625 

6671 

6717 

6763 

46 

8 

6808 

6854 

6900 

6946 

6992 

7037 

7083 

7129 

7175 

7220 

46 

9 

7266 

7312 

7358 

7403 

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7495 

7541 

7586 

7632 

7678 

46 

950 

977724 

977769 

977815 

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978043 

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978135 

46 

1 

8181 

8226 

8272 

8317 

8363 

8409 

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8500 

8546 

8591 

46 

2 

8637 

8683 

8728 

8774 

8819 

8865 

8911 

8950 

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9047 

46 

3 

9093 

9138 

9184 

9230 

9275 

9-321 

9366 

9412 

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46 

4 

9548 

9594 

9639 

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9730 

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9821 

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46 

5 

980003 

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980140 

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980231 

980276 

980322 

980367 

980412 

45 

G 

0458 

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0549 

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45 

7 

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1003 

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1139 

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1229 

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45 

8 

1366 

1411 

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1501 

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1637 

1683 

1728 

1773 

45 

9 

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1864 

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1954 

2000 

2045 

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45 

1 

2723 

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2994 

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45 

2 

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45 

3 

3626 

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45 

4 

4077 

4122 

4167 

4212 

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4437 

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5 

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45 

6 

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5022 

5067 

5112 

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5202 

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5292 

5337 

5382 

45 

7 

5426 

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45 

8 

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45 

1  ^ 

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